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in Figure 16.25 to illustrate how the ray changes Access for free at openstax.org. direction both as it enters and as it leaves the lens. Because the index of refraction of the lens is greater than that of air, the ray moves toward the perpendicular as it enters and away from the perpendicular as it leaves. (This is in accordance with the law of refraction.) As a result of the shape of the lens, light is thus bent toward the axis at both surfaces. 16.3 • Lenses 499 Figure 16.25 Rays of light entering a convex, or converging, lens parallel to its axis converge at its focal point, F. Ray 2 lies on the axis of the lens. The distance from the center of the lens to the focal point is the focal length, ƒ, of the lens. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. Note that rays from a light source placed at the focal point of a converging lens emerge parallel from the other side of the lens. You may have heard of the trick of using a converging lens to focus rays of sunlight to a point. Such a concentration of light energy can produce enough heat to ignite paper. Figure 16.26 shows a concave lens and the effect it has on rays of light that enter it parallel to its axis (the path taken by ray 2 in the figure is the axis of the lens). The concave lens is a diverging lens because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens. The distance from the center of the lens to the focal point is again called the focal length, or “ƒ,” of the lens. Note that the focal length of a diverging lens is defined to be negative. An expanded view of the path of one ray through the lens is shown in Figure 16.26 to illustrate how the shape of the lens, together with the law of refraction, causes the ray to follow its particular path and diverge. Figure 16.26 Rays of light enter a concave, or diverging, lens parallel to its axis diverge and thus appear to originate from its focal point, F. The dashed lines are not rays; they indicate the directions
from which the rays appear to come. The focal length, ƒ, of a diverging lens is negative. An expanded view of the path taken by ray 1 shows the perpendiculars and the angles of incidence and refraction at both surfaces. The power, P, of a lens is very easy to calculate. It is simply the reciprocal of the focal length, expressed in meters 16.15 The units of power are diopters, D, which are expressed in reciprocal meters. If the focal length is negative, as it is for the diverging lens in Figure 16.26, then the power is also negative. In some circumstances, a lens forms an image at an obvious location, such as when a movie projector casts an image onto a screen. In other cases, the image location is less obvious. Where, for example, is the image formed by eyeglasses? We use ray 500 Chapter 16 • Mirrors and Lenses tracing for thin lenses to illustrate how they form images, and we develop equations to describe the image-formation quantitatively. These are the rules for ray tracing: 1. A ray entering a converging lens parallel to its axis passes through the focal point, F, of the lens on the other side 2. A ray entering a diverging lens parallel to its axis seems to come from the focal point, F, on the side of the entering ray 3. A ray passing through the center of either a converging or a diverging lens does not change direction 4. A ray entering a converging lens through its focal point exits parallel to its axis 5. A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to the axis Consider an object some distance away from a converging lens, as shown in Figure 16.27. To find the location and size of the image formed, we trace the paths of select light rays originating from one point on the object. In this example, the originating point is the top of a woman’s head. Figure 16.27 shows three rays from the top of the object that can be traced using the raytracing rules just listed. Rays leave this point traveling in many directions, but we concentrate on only a few, which have paths that are easy to trace. The first ray is one that enters the lens parallel to its axis and passes through the focal point on the other side (rule 1). The second ray passes through the center of the lens without changing direction (rule 3). The third ray passes through
the nearer focal point on its way into the lens and leaves the lens parallel to its axis (rule 4). All rays that come from the same point on the top of the person’s head are refracted in such a way as to cross at the same point on the other side of the lens. The image of the top of the person’s head is located at this point. Rays from another point on the object, such as the belt buckle, also cross at another common point, forming a complete image, as shown. Although three rays are traced in Figure 16.27, only two are necessary to locate the image. It is best to trace rays for which there are simple ray-tracing rules. Before applying ray tracing to other situations, let us consider the example shown in Figure 16.27 in more detail. Access for free at openstax.org. 16.3 • Lenses 501 Figure 16.27 Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced. The three chosen rays each follow one of the rules for ray tracing, so their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed. The image formed in Figure 16.27 is a real image—meaning, it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye. In Figure 16.27, the object distance, do, is greater than f.Now we consider a ray diagram for a convex lens where do<f, and another diagram for a concave lens. 502 Chapter 16 • Mirrors and Lenses Virtual Physics Geometric Optics Click to view content (https://www.openstax.org/l/28Geometric) This animation shows you how the image formed by a convex lens changes as you change object distance, curvature radius, refractive index, and diameter of the lens. To begin, choose Principal Rays in the upper left menu and then try varying some of the parameters indicated at the upper center. Show Help supplies a few helpful labels., change with an increasing radius of curvature? How does change with an increasing How does the focal length, refractive index? a. The focal length increases in both cases: when the
radius of curvature and the refractive index increase. b. The focal length decreases in both cases: when the radius of curvature and the refractive index increase. c. The focal length increases when the radius of curvature increases; it decreases when the refractive index increases. d. The focal length decreases when the radius of curvature increases; it increases in when the refractive index increases. Type Formed When Image Type di M Case 1 fpositive, do> f Real Positive Negative m>, <, or = ‒1 Case 2 fpositive, do< f Virtual Negative Positive m> 1 Case 3 fnegative Virtual Negative Positive m< 1 Table 16.3 Three Types of Images Formed by Lenses The examples in Figure 16.27 and Figure 16.28 represent the three possible cases—case 1, case 2, and case 3—summarized in Table 16.3. In the table, mis magnification; the other symbols have the same meaning as they did for curved mirrors. Figure 16.28 (a) The image is virtual and larger than the object. (b) The image is virtual and smaller than the object. Access for free at openstax.org. 16.3 • Lenses 503 Snap Lab Focal Length • Temperature extremes—Very hot or very cold temperatures are encountered in this lab that can cause burns. Use protective mitts, eyewear, and clothing when handling very hot or very cold objects. Notify your teacher immediately of any burns. • EYE SAFETY—Looking at the Sun directly can cause permanent eye damage. Do not look at the Sun through any lens. • Several lenses • A sheet of white paper • A ruler or tape measure Instructions Procedure 1. Find several lenses and determine whether they are converging or diverging. In general, those that are thicker near the edges are diverging and those that are thicker near the center are converging. 2. On a bright, sunny day take the converging lenses outside and try focusing the sunlight onto a sheet of white paper. 3. Determine the focal lengths of the lenses. Have one partner slowly move the lens toward and away from the paper until you find the distance at which the light spot is at its brightest. Have the other partner measure the distance from the lens to the bright spot. Be careful, because the paper may start to burn, depending on the type of lens. True or false—The bright spot that appears in focus on the paper is an image of the Sun.
a. True b. False Image formation by lenses can also be calculated from simple equations. We learn how these calculations are carried out near the end of this section. Some common applications of lenses with which we are all familiar are magnifying glasses, eyeglasses, cameras, microscopes, and telescopes. We take a look at the latter two examples, which are the most complex. We have already seen the design of a telescope that uses only mirrors in. Figure 16.29 shows the design of a telescope that uses two lenses. Part (a) of the figure shows the design of the telescope used by Galileo. It produces an upright image, which is more convenient for many applications. Part (b) shows an arrangement of lenses used in many astronomical telescopes. This design produces an inverted image, which is less of a problem when viewing celestial objects. 504 Chapter 16 • Mirrors and Lenses Figure 16.29 (a) Galileo made telescopes with a convex objective and a concave eyepiece. They produce an upright image and are used in spyglasses. (b) Most simple telescopes have two convex lenses. The objective forms a case 1 image, which is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified. Figure 16.30 shows the path of light through a typical microscope. Microscopes were first developed during the early 1600s by eyeglass makers in the Netherlands and Denmark. The simplest compound microscope is constructed from two convex lenses, as shown schematically in Figure 16.30. The first lens is called the objective lens; it has typical magnification values from 5 to 100. In standard microscopes, the objectives are mounted such that when you switch between them, the sample remains in focus. Objectives arranged in this way are described as parfocal. The second lens, the eyepiece, also referred to as the ocular, has several lenses that slide inside a cylindrical barrel. The focusing ability is provided by the movement of both the objective lens and the eyepiece. The purpose of a microscope is to magnify small objects, and both lenses contribute to the final magnification. In addition, the final enlarged image is produced in a location far enough from the observer to be viewed easily because the eye cannot focus on objects or images that are too close. Access for free at openstax.org. 16.3 • Lenses 505 Figure 16.30 A compound microscope composed of two lenses, an objective and an
eyepiece. The objective forms a case 1 image that is larger than the object. This first image is the object for the eyepiece. The eyepiece forms a case 2 final image that is magnified even further. Real lenses behave somewhat differently from how they are modeled using rays diagrams or the thin-lens equations. Real lenses produce aberrations. An aberration is a distortion in an image. There are a variety of aberrations that result from lens size, material, thickness, and the position of the object. One common type of aberration is chromatic aberration, which is related to color. Because the index of refraction of lenses depends on color, or wavelength, images are produced at different places and with different magnifications for different colors. The law of reflection is independent of wavelength, so mirrors do not have this problem. This result is another advantage for the use of mirrors in optical systems such as telescopes. Figure 16.31(a) shows chromatic aberration for a single convex lens, and its partial correction with a two-lens system. The index of refraction of the lens increases with decreasing wavelength, so violet rays are refracted more than red rays, and are thus focused closer to the lens. The diverging lens corrects this in part, although it is usually not possible to do so completely. Lenses made of different materials and with different dispersions may be used. For example, an achromatic doublet consisting of a converging lens made of crown glass in contact with a diverging lens made of flint glass can reduce chromatic aberration dramatically (Figure 16.31(b)). Figure 16.31 (a) Chromatic aberration is caused by the dependence of a lens’s index of refraction on color (wavelength). The lens is more powerful for violet (V) than for red (R), producing images with different colors, locations, and magnifications. (b) Multiple-lens systems can correct chromatic aberrations in part, but they may require lenses of different materials and add to the expense of optical systems such as cameras. 506 Chapter 16 • Mirrors and Lenses Physics of the Eye The eye is perhaps the most interesting of all optical instruments. It is remarkable in how it forms images and in the richness of detail and color they eye can detect. However, our eyes commonly need some correction to reach what is called normalvision, but should be called idealvision instead. Image formation by our
ly while traveling from air into the cornea. The lens provides the remaining magnification needed to produce an image on the retina. The cornea and lens can be treated as a single thin lens, although the light rays pass through several layers of material (such as the cornea, aqueous humor, several layers in the lens, and vitreous humor), changing direction at each interface. The image formed is much like the one produced by a single convex lens. This result is a case 1 image. Images formed in the eye are inverted, but the brain inverts them once more to make them seem upright. Material Index of Refraction Water Air Cornea Aqueous humor 1.33 1.00 1.38 1.34 The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Access for free at openstax.org. 16.3 • Lenses 507 Material Index of Refraction Lens 1.41 average Vitreous humor 1.34 *The index of refraction varies throughout the lens and is greatest at its center. Table 16.4 Refractive Indices Relevant to the Eye Figure 16.33 An image is formed on the retina, with light rays converging most at the cornea and on entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale. As noted, the image must fall precisely on the retina to produce clear vision—that is, the image distance, di, must equal the lensto-retina distance. Because the lens-to-retina distance does not change, di must be the same for objects at all distances. The eye manages to vary the distance by varying the power (and focal length) of the lens to accommodate for objects at various distances. In Figure 16.33, you can see the small ciliary muscles above and below the lens that change the shape of the lens and, thus, the focal length. The need for some type of vision correction is very common. Common vision defects are easy to understand, and some are simple to correct. Figure 16.34 illustrates two common vision defects. Nearsightedness, or myopia, is the inability to see distant objects clearly while close objects are in focus. The nearsighted eye overconvergesthe nearly parallel rays from a distant object, and the rays
cross in front of the retina. More divergent rays from a close object are converged on the retina, producing a clear image. Farsightedness, or hyperopia, is the inability to see close objects clearly whereas distant objects may be in focus. A farsighted eye does not converge rays from a close object sufficiently to make the rays meet on the retina. Less divergent rays from a distant object can be converged for a clear image. Figure 16.34 (a) The nearsighted (myopic) eye converges rays from a distant object in front of the retina; thus, they are diverging when they strike the retina, and produce a blurry image. This divergence can be caused by the lens of the eye being too powerful (in other words, too short a focal length) or the length of the eye being too great. (b) The farsighted (hyperopic) eye is unable to converge the rays from a close object by the time they strike the retina and produce... blurry close vision. This poor convergence can be caused by insufficient power (in 508 Chapter 16 • Mirrors and Lenses other words, too long a focal length) in the lens or by the eye being too short. Because the nearsighted eye overconverges light rays, the correction for nearsightedness involves placing a diverging spectacle lens in front of the eye. This lens reduces the power of an eye that has too short a focal length (Figure 16.35(a)). Because the farsighted eye underconvergeslight rays, the correction for farsightedness is to place a converging spectacle lens in front of the eye. This lens increases the power of an eye that has too long a focal length (Figure 16.35(b)). Figure 16.35 (a) Correction of nearsightedness requires a diverging lens that compensates for the overconvergence by the eye. The diverging lens produces an image closer to the eye than the object so that the nearsighted person can see it clearly. (b) Correction of farsightedness uses a converging lens that compensates for the underconvergence by the eye. The converging lens produces an image farther from the eye than the object so that the farsighted person can see it clearly. In both (a) and (b), the rays that meet at the retina represent corrected vision, and the other rays represent blurred vision without corrective lenses. Calculations Using Lens
Equations As promised, there are no new equations to memorize. We can use equations already presented for solving problems involving curved mirrors. Careful analysis allows you to apply these equations to lenses. Here are the equations you need where Pis power, expressed in reciprocal meters (m–1) rather than diopters (D), and fis focal length, expressed in meters (m). You also need where, as before, do and di are object distance and image distance, respectively. Remember, this equation is usually more useful if rearranged to solve for one of the variables. For example, The equations for magnification, m, are also the same as for mirrors where hi and ho are the image height and object height, respectively. Remember, also, that a negative di value indicates a virtual image and a negative hi value indicates an inverted image. These are the steps to follow when solving a lens problem: • Step 1. Examine the situation to determine that image formation by a lens is involved. • Step 2. Determine whether ray tracing, the thin-lens equations, or both should be used. A sketch is very helpful even if ray tracing is not specifically required by the problem. Write useful symbols and values on the sketch. • Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). • Step 4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is helpful to determine whether the situation involves a case 1, 2, or 3 image. Although these are just names for types of images, they have certain characteristics (given in Table 16.3) that can be of great use in solving problems. Access for free at openstax.org. 16.3 • Lenses 509 • Step 5. If ray tracing is required, use the ray-tracing rules listed earlier in this section. • Step 6. Most quantitative problems require the use of the thin-lens equations. These equations are solved in the usual manner by substituting knowns and solving for unknowns. Several worked examples were included earlier and can serve as guides. • Step 7. Check whether the answer is reasonable. Does it make sense?If you identified the type of image (case 1, 2, or 3) correctly, you should assess whether your answer is consistent with the type of image, magnification, and so on. All problems will be solved by one or more of the equations just presented, with
ray tracing used only for general analysis of the problem. The steps then simplify to the following: Identify the unknown. Identify the knowns. 1. 2. 3. Choose an equation, plug in the knowns, and solve for the unknown. Here are some worked examples: WORKED EXAMPLE The Power of a Magnifying Glass Strategy The Sun is so far away that its rays are nearly parallel when they reach Earth. The magnifying glass is a convex (or converging) lens, focusing the nearly parallel rays of sunlight. Thus, the focal length of the lens is the distance from the lens to the spot, and its power, in diopters (D), is the inverse of this distance (in reciprocal meters). Solution The focal length of the lens is the distance from the center of the lens to the spot, which we know to be 8.00 cm. Thus, 16.16 To find the power of the lens, we must first convert the focal length to meters; then, we substitute this value into the equation for power. Discussion This result demonstrates a relatively powerful lens. Remember that the power of a lens in diopters should not be confused with the familiar concept of power in watts. 16.17 WORKED EXAMPLE Image Formation by a Convex Lens A clear glass light bulb is placed 0.75 m from a convex lens with a 0.50 m focal length, as shown in Figure 16.36. Use ray tracing to get an approximate location for the image. Then, use the mirror/lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin-lens and magnification equations produce consistent results. Figure 16.36 A light bulb placed 0.75 m from a lens with a 0.50 m focal length produces a real image on a poster board, as discussed in the previous example. Ray tracing predicts the image location and size. 510 Chapter 16 • Mirrors and Lenses Strategy Because the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to the one illustrated in the previous figure of a series of drawings showing a woman standing to the left of a lens. Ray tracing to scale should produce similar results for di. Numerical solutions for di and mcan be obtained using the thin-lens and magnification equations, noting that do = 0.75 m and f= 0.50 m. Solution The ray tracing
to scale in Figure 16.36 shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus, the image distance, di, is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of two, and the image is inverted. Thus, mis about –2. The minus sign indicates the image is inverted. The lens equation can be rearranged to solve for di from the given information. Now, we use to find m. 16.18 16.19 Discussion Note that the minus sign causes the magnification to be negative when the image is inverted. Ray tracing and the use of the lens equation produce consistent results. The thin-lens equation gives the most precise results, and is limited only by the accuracy of the given information. Ray tracing is limited by the accuracy with which you draw, but it is highly useful both conceptually and visually. WORKED EXAMPLE Image Formation by a Concave Lens Suppose an object, such as a book page, is held 6.50 cm from a concave lens with a focal length of –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced? Strategy This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is therefore the same, but the results are different in important ways. Solution 16.20 Now the magnification equation can be used to find the magnification, m, because both di and do are known. Entering their values gives 16.21 Discussion A number of results in this example are true of all case 3 images. Magnification is positive (as calculated), meaning the image is upright. The magnification is also less than one, meaning the image is smaller than the object—in this case, a little more than half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. The image is virtual. The image is closer to the lens than the object, because the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, because the image is smaller than the object, you may think it is farther away; however, the image is closer than the object—a fact that is useful in correcting nearsighted
ness. Access for free at openstax.org. 16.3 • Lenses 511 WATCH PHYSICS The Lens Equation and Problem Solving The video shows calculations for both concave and convex lenses. It also explains real versus virtual images, erect versus inverted images, and the significance of negative and positive signs for the involved variables. Click to view content (https://www.openstax.org/l/28Lenses) If a lens has a magnification of a. The image is erect and is half as tall as the object. b. The image is erect and twice as tall as the object. c. The image is inverted and is half as tall as the object. d. The image is inverted and is twice as tall as the object., how does the image compare with the object in height and orientation? Practice Problems 11. A lens has a focal length of. What is the power of the lens? a. The power of the lens is b. The power of the lens is c. The power of the lens is d. The power of the lens is.... 12. If a lens produces a 5.00 -cm tall image of an 8.00 -cm -high object when placed 10.0 cm from the lens, what is the apparent image distance? Construct a ray diagram using paper, a pencil, and a ruler to confirm your calculation. a. −3.12 cm b. −6.25 cm 3.12 cm c. d. 6.25 cm Check Your Understanding 13. A lens has a magnification that is negative. What is the orientation of the image? a. Negative magnification means the image is erect and real. b. Negative magnification means the image is erect and virtual. c. Negative magnification means the image is inverted and virtual. d. Negative magnification means the image is inverted and real. 14. Which part of the eye controls the amount of light that enters? a. b. c. d. the pupil the iris the cornea the retina 15. An object is placed between the focal point and a convex lens. Describe the image that is formed in terms of its orientation, and whether the image is real or virtual. a. The image is real and erect. b. The image is real and inverted. c. The image is virtual and erect. d. The image is virtual and inverted. 16. A farsighted person buys a pair of glasses to correct her farsightedness
. Describe the main symptom of farsightedness and the type of lens that corrects it. a. Farsighted people cannot focus on objects that are far away, but they can see nearby objects easily. A convex lens is used to correct this. b. Farsighted people cannot focus on objects that are close up, but they can see far-off objects easily. A concave lens is used to correct this. 512 Chapter 16 • Mirrors and Lenses c. Farsighted people cannot focus on objects that are close up, but they can see distant objects easily. A convex lens is used to correct this. d. Farsighted people cannot focus on objects that are either close up or far away. A concave lens is used to correct this. Access for free at openstax.org. Chapter 16 • Key Terms 513 KEY TERMS aberration a distortion in an image produced by a lens angle of incidence the angle, with respect to the normal, at which a ray meets a boundary between media or a reflective surface focal point the point at which rays converge or appear to converge incident ray the incoming ray toward a medium boundary or a reflective surface angle of reflection the angle, with respect to the normal, at index of refraction the speed of light in a vacuum divided which a ray leaves a reflective surface by the speed of light in a given material angle of refraction the angle between the normal and the law of reflection the law that indicates the angle of refracted ray reflection equals the angle of incidence central axis a line perpendicular to the center of a lens or law of refraction the law that describes the relationship mirror extending in both directions chromatic aberration an aberration related to color concave lens a lens that causes light rays to diverge from the central axis concave mirror a mirror with a reflective side that is curved inward converging lens a convex lens convex lens a lens that causes light rays to converge toward the central axis between refractive indices of materials on both sides of a boundary and the change in the path of light crossing the boundary, as given by the equation n1 sin = n2 sin ray light traveling in a straight line real image an optical image formed when light rays converge and pass through the image, producing an image that can be projected onto a screen refracted ray the light ray after it has been refracted Snell’s law the law of refraction expressed mathematically convex mirror a mirror with a reflective side that is
curved as outward critical angle an incident angle that produces an angle of refraction of 90° dispersion separation of white light into its component wavelengths diverging lens a concave lens focal length the distance from the focal point to the mirror SECTION SUMMARY 16.1 Reflection • The angle of reflection equals the angle of incidence. • Plane mirrors and convex mirrors reflect virtual, erect images. Concave mirrors reflect light to form real, inverted images or virtual, erect images, depending on the location of the object. Image distance, height, and other characteristics can be calculated using the lens/mirror equation and the magnification equation. • 16.2 Refraction • The index of refraction for a material is given by the speed of light in a vacuum divided by the speed of light in that material. • Snell’s law states the relationship between indices of KEY EQUATIONS 16.1 Reflection lens/mirror equation (reciprocal version) total internal reflection reflection of light traveling through a medium with a large refractive index at a boundary of a medium with a low refractive index under conditions such that refraction cannot occur virtual image the point from which light rays appear to diverge without actually doing so refraction, the incident angle, and the angle of refraction. • The critical angle,, determines whether total internal refraction can take place, and can be calculated according to. 16.3 Lenses • The characteristics of images formed by concave and convex lenses can be predicted using ray tracing. Characteristics include real versus virtual, inverted versus upright, and size. • The human eye and corrective lenses can be explained using geometric optics. • Characteristics of images formed by lenses can be calculated using the mirror/lens equation. lens/mirror equation (solved version) 514 Chapter 16 • Chapter Review magnification equation critical angle radius/focal length equation R= 2f 16.2 Refraction index of Refraction Snell’s law >Snell’s law in terms of speed CHAPTER REVIEW Concept Items 16.1 Reflection 1. Part A. Can you see a virtual image? Part B. Can you photograph one? Explain your answers. a. A. yes; B. No, an image from a flat mirror cannot be photographed. b. A. no; B. Yes, an image from a flat mirror can be photographed. c. A. yes; B. Yes, an image from a flat mirror can be photographed. d. A.
no; B. No, an image from a flat mirror cannot be photographed. 2. State the law of reflection. a. b. c. d., where is the angle of incidence., where is the angle of incidence., where is the angle of incidence., where is the angle of reflection and is the angle of reflection and is the angle of reflection and is the angle of reflection. 16.2 Refraction 3. Does light change direction toward or away from the normal when it goes from air to water? Explain. a. The light bends away from the normal because the index of refraction of water is greater than that of air. b. The light bends away from the normal because the index of refraction of air is greater than that of water. Access for free at openstax.org. 16.3 Lenses power and focal length mirror/lens (or thin-lens) equation rearranged mirror/lens equation magnification equation c. The light bends toward the normal because the index of refraction of water is greater than that of air. d. The light bends toward the normal because the index of refraction of air is greater than that of water. 16.3 Lenses 4. An object is positioned in front of a lens with its base resting on the principal axis. Describe two rays that could be traced from the top of the object and through the lens that would locate the top of an image. a. A ray perpendicular to the axis and a ray through the center of the lens b. A ray parallel to the axis and a ray that does not pass through the center of the lens c. A ray parallel to the axis and a ray through the center of the lens d. A ray parallel to the axis and a ray that does not pass through the focal point 5. A person timing the moonrise looks at her watch and then at the rising moon. Describe what happened inside her eyes that allowed her to see her watch clearly one second and then see the moon clearly. a. The shape of the lens was changed by the sclera, and thus its focal length was also changed, so that each of the images focused on the retina. b. The shape of the lens was changed by the choroid, and thus its focal length was also changed, so that each of the images focused on the retina. c. The shape of the lens was changed by the iris, and thus its focal length was also changed, so that each of the
images focused on the retina. d. The shape of the lens was changed by the muscles, and thus its focal length was also changed, so that each of the images focused on the retina. 6. For a concave lens, if the image distance, di, is negative, where does the image appear to be with respect to the object? a. The image always appears on the same side of the Critical Thinking Items 16.1 Reflection 7. Why are diverging mirrors often used for rear-view mirrors in vehicles? What is the main disadvantage of using such a mirror compared with a flat one? a. It gives a wide range of view. The image appears to be closer than the actual object. It gives a narrow range of view. The image appears to be farther than the actual object. It gives a narrow range of view. The image appears to be closer than the actual object. It gives a wide range of view. The image appears to be farther than the actual object. b. c. d. 16.2 Refraction 8. A high-quality diamond may be quite clear and colorless, transmitting all visible wavelengths with little absorption. Explain how it can sparkle with flashes of brilliant color when illuminated by white light. a. Diamond and air have a small difference in their refractive indices that results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. b. Diamond and air have a small difference in their refractive indices that results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. c. Diamond has a high index of refraction with respect to air, which results in a very small critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. d. Diamond has a high index of refraction with respect to air, which results in a very large critical angle. The light that enters a diamond may exit at only a few points, and these points sparkle because many rays have been directed toward them. 9. The most common type of mirage is an illusion in which light from far-away objects is reflected by a pool of water that is not really there. Mirages are generally observed in Chapter 16 • Chapter Review 515 lens. b. The image appears
on the opposite side of the lens. c. The image appears on the opposite side of the lens only if the object distance is greater than the focal length. d. The image appears on the same side of the lens only if the object distance is less than the focal length. deserts, where there is a hot layer of air near the ground. Given that the refractive index of air is less for air at higher temperatures, explain how mirages can be formed. a. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. b. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in refractive index is large, which results in a large critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. c. The hot layer of air near the ground is lighter than the cooler air above it, but the difference in refractive index is small, which results in a small critical angle. The light rays coming from the horizon strike the hot air at large angles, so they are reflected as they would be from water. d. The hot layer of air near the ground is lighter than the cooler air above it, and the difference in the refractive index is large, which results in a small critical angle. The light rays coming from the horizontal strike the hot air at large angles, so they are reflected as they would be from water. 16.3 Lenses 10. When you focus a camera, you adjust the distance of the lens from the film. If the camera lens acts like a thin lens, why can it not be kept at a fixed distance from the film for both near and distant objects? a. To focus on a distant object, you need to increase the image distance. b. To focus on a distant object, you need to increase the focal length of the lens. c. To focus on a distant object, you need to decrease the focal length of the lens. d. To focus on a distant object, you may need to increase or decrease the focal length of the lens. 11. Part A—How do the refractive indices of the cornea, 516 Chapter 16 • Chapter Review aqueous humor, and the
lens of the eye compare with the refractive index of air? Part B—How do the comparisons in part A explain how images are focused on the retina? a. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted away from the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted away from b. Problems 16.1 Reflection 12. Some telephoto cameras use a mirror rather than a lens. What radius of curvature is needed for a concave mirror to replace a 0.800 -m focal-length telephoto lens? a. 0.400 m 1.60 m b. c. 4.00 m 16.0 m d. 13. What is the focal length of a makeup mirror that produces a magnification of 2.00 when a person’s face is 8.00 cm away? a. –16 cm b. –5.3 cm c. d. 5.3 cm 16 cm c. d. the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have smaller refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. (A) The cornea, aqueous humor, and lens of the eye have greater refractive indices than air. (B) Rays entering the eye are refracted toward the central axis, which causes them to meet at the focal point on the retina. Calculate the amount the ray is displaced by the glass (Δx), given that the incident angle is 40.0° and the glass is 1.00 cm thick. a. 0.839 cm b. 0.619 cm c. 0.466 cm d. 0.373 cm 16.2 Refraction 14. An optical fiber uses flint glass (n= 1.66) clad with crown 16.3 Lenses glass (n= 1.52). What is the critical angle? a. 33.2° b. 23.7° c. 0.92 rad 1.16 rad d. 15
. Suppose this figure represents a ray of light going from air (n= 1.0003) through crown glass (n= 1.52) into water, similar to a beam of light going into a fish tank. 16. A camera’s zoom lens has an adjustable focal length ranging from 80.0 to 200 mm. What is its range of powers? a. The lowest power is 0.05 D and the highest power is 0.125 D. b. The lowest power is 0.08 D and the highest power is 0.20 D. c. The lowest power is 5.00 D and the highest power is 12.5 D. d. The lowest power is 80 D and the highest power is 200 D. 17. Suppose a telephoto lens with a focal length of 200 mm is being used to photograph mountains 10.0 km away. (a) Where is the image? (b) What is the height of the image of a 1,000-m-high cliff on one of the mountains? a. (a) The image is 0.200 m on the same side of the lens. (b) The height of the image is – 2.00 cm. (a) The image is 0.200 m on the opposite side of the Access for free at openstax.org. b. Chapter 16 • Test Prep 517 c. lens. (b) The height of the image is – 2.00 cm. (a) The image is 0.200 m on the opposite side of the lens. (b) The height of the image is +2.00 cm. d. (a) The image is 0.100 m on the same side of the lens. (b) The height of the image is +2.00 cm. Performance Task 16.3 Lenses 18. In this performance task, you will investigate the lens- like properties of a clear bottle. • a water bottle or glass with a round cross-section and smooth, vertical sides • enough water to fill the bottle • a meter stick or tape measure • a bright light source with a small bulb, such as a pen light • a small bright object, such as a silver spoon. Instructions TEST PREP Multiple Choice 16.1 Reflection 19. In geometric optics, a straight line emerging from a point is called a (an) ________. a. b. c. d. object distance ray focal point image 20. An image of a 2.0 -cm
object reflected from a mirror is 5.0 cm tall. What is the magnification of the mirror? a. 0.4 b. 2.5 3 c. 10 d. 21. Can a virtual image be projected onto a screen with additional lenses or mirrors? Explain your answer. a. Yes, the rays actually meet behind the lens or mirror. b. No, the image is formed by rays that converge to a point in front of the mirror or lens. Procedure 1. Look through a clear glass or plastic bottle and describe what you see. 2. Next, fill the bottle with water and describe what you see. 3. Use the water bottle as a lens to produce the image of a bright object. 4. Estimate the focal length of the water bottle lens. a. How can you find the focal length of the lens using the light and a blank wall? b. How can you find the focal length of the lens using the bright object? c. Why did the water change the lens properties of the bottle? b. c. d. the refractive index the speed of light in a vacuum the speed of light in a transparent material 23. What is the term for the minimum angle at which a light ray is reflected back into a material and cannot pass into the surrounding medium? critical angle a. b. incident angle c. angle of refraction d. angle of reflection 24. Consider these indices of refraction: glass: 1.52, air: 1.0003, water: 1.333. Put these materials in order from the one in which the speed of light is fastest to the one in which it is slowest. a. The speed of light in water > the speed of light in air > the speed of light in glass. b. The speed of light in glass > the speed of light in water > the speed of light in air. c. The speed of light in air > the speed of light in water > the speed of light in glass. d. The speed of light in glass > the speed of light in air c. Yes, any image that can be seen can be manipulated > the speed of light in water. so that it can be projected onto a screen. d. No, the image can only be perceived as being behind the lens or mirror. 16.2 Refraction 22. What does crepresent in the equation? a. the critical angle 25. Explain why an object in water always appears to be at a depth that is more shallow than it actually
is. a. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to 518 Chapter 16 • Test Prep be at a depth that is more shallow than the actual depth. b. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is above the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. c. Because of the refraction of light, the light coming from the object bends toward the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. d. Because of the refraction of light, the light coming from the object bends away from the normal at the interface of water and air. This causes the object to appear at a location that is below the actual position of the object. Hence, the image appears to be at a depth that is more shallow than the actual depth. 16.3 Lenses 26. For a given lens, what is the height of the image divided by the height of the object ( ) equal to? a. power focal length b. c. magnification d. radius of curvature Short Answer 16.1 Reflection 30. Distinguish between reflection and refraction in terms of how a light ray changes when it meets the interface between two media. a. Reflected light penetrates the surface whereas refracted light is bentas it travels from one medium to the other. b. Reflected light penetrates the surface whereas refracted light travels along a curved path. c. Reflected light bouncesfrom the surface whereas refracted light travels along a curved path. d. Reflected light bouncesfrom the surface whereas refracted light is bentas it travels from one medium to the other. 31. Sometimes light may be both reflected and refracted as it meets the surface of a different medium. Identify a material with a surface that when light travels through Access for free at openstax.org. 27. Which part of the eye has the greatest density of light receptors? a. b. c. d. the
lens the fovea the optic nerve the vitreous humor 28. What is the power of a lens with a focal length of 10 cm? a. b. c. d. 10 m–1, or 10 D 10 cm–1, or 10 D 10 m, or 10 D 10 cm, or 10 D 29. Describe the cause of chromatic aberration. a. Chromatic aberration results from the dependence of the frequency of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. b. Chromatic aberration results from the dispersion of different wavelengths of light by a curved mirror so that each color has a different focal point. c. Chromatic aberration results from the dependence of the reflection angle at a spherical mirror’s surface on the distance of light rays from the principal axis so that different colors have different focal points. d. Chromatic aberration results from the dependence of the wavelength of light on the refractive index, which causes dispersion of different colors of light by a lens so that each color has a different focal point. the air it is both reflected and refracted. Explain how this is possible. a. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the transparent glass. b. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface, and it is refracted while passing into the opaque glass. c. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a smooth surface; it is refracted while passing into the opaque glass. d. Light passing through air is partially reflected and refracted when it meets a glass surface. It is reflected because glass has a rough surface; it is refracted while passing into the transparent glass. 32. A concave mirror has a focal length of 5.00 cm. What is the image distance of an object placed 7.00 cm from the center of the mirror? a. −17.5 cm b. −2.92 cm c. 2.92 cm 17.5 cm d. 33. An 8.0 -cm tall object is placed 6.0 cm from a concave mirror with a magnification of –2.0. What are the image height and the
image distance? a. hi = – 16 cm, di = – 12 cm b. hi = – 16 cm, di = 12 cm c. hi = 16 cm, di = – 12 cm d. hi = 16 cm, di = 12 cm 16.2 Refraction 34. At what minimum angle does total internal reflection of toward light occur if it travels from water ice? a. b. c. d. 35. Water floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0° and an angle of refraction of 40.1°. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride? a. b. c. d. 1.60 1.49 1.21 1.46 36. Describe what happens to a light ray when it is refracted. Include in your explanation comparison of angles, comparison of refractive indices, and the term normal. a. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of interference. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is greater in the medium with the greater refractive index. b. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the greater refractive index. Chapter 16 • Test Prep 519 another medium with a different refractive index, the ray does not change its path. The angle between the ray and the normal (the line parallel to the surfaces of the two media) is the same in both media. d. When a ray of light goes from one medium to another medium with a different refractive index, the ray changes its path as a result of refraction. The angle between the ray and the normal (the line perpendicular to the surfaces of the two media) is less in the medium with the lower refractive index. 16.3 Lenses 37. What are two equivalent terms for a lens that always causes light rays to bend away from the principal axis? a. a diverging lens or a convex lens b. a
diverging lens or a concave lens c. a converging lens or a concave lens d. a converging lens or a convex lens 38. Define the term virtual image. a. A virtual image is an image that cannot be projected onto a screen. b. A virtual image is an image that can be projected onto a screen. c. A virtual image is an image that is formed on the opposite side of the lens from where the object is placed. d. A virtual image is an image that is always bigger than the object. 39. Compare nearsightedness (myopia) and farsightedness (hyperopia) in terms of focal point. a. The eyes of a nearsighted person have focal points beyond the retina. A farsighted person has eyes with focal points between the lens and the retina. b. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points beyond the retina. c. A nearsighted person has eyes with focal points between the lens and the choroid. A farsighted person has eyes with focal points beyond the choroid. d. A nearsighted person has eyes with focal points between the lens and the retina. A farsighted person has eyes with focal points on the retina. 40. Explain how a converging lens corrects farsightedness. a. A converging lens disperses the rays so they focus on the retina. b. A converging lens bends the rays closer together so they do not focus on the retina. c. A converging lens bends the rays closer together so c. When a ray of light goes from one medium to they focus on the retina. 520 Chapter 16 • Test Prep d. A converging lens disperses the rays so they do not 42. What is the magnification of a lens if it produces a focus on the retina. 41. Solve the equation for in such a way that it is not expressed as a reciprocal. 12-cm-high image of a 4 -cm -high object? The image is virtual and erect. a. b. a. b. c. d. c. d. Extended Response 16.1 Reflection 43. The diagram shows a lightbulb between two mirrors. One mirror produces a beam of light with parallel rays; the other keeps light from escaping without being put into the beam. What angle does the ray make from the instructor’s face
with the normal to the water (n= 1.33) at the point where the ray enters? Assume n= 1.00 for air. a. 68° b. 25° 19° c. 34° d. 46. Describe total internal reflection. Include a definition of the critical angle and how it is related to total internal reflection. Also, compare the indices of refraction of the interior material and the surrounding material. a. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90°. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. b. When the interior material has a smaller index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is less than 90°. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. c. When the interior material has the same index of refraction as the surrounding material, the incident ray approaches the boundary at an angle (called the critical angle) such that the refraction Where is the light source in relation to the focal point or radius of curvature of each mirror? Explain your answer. a. The bulb is at the center of curvature of the small mirror and at the focal point of the large mirror. b. The bulb is at the focal point of the small mirror and at the focal point of the large mirror. c. The bulb is at the center of curvature of the small mirror and at the center of curvature of the large mirror. d. The bulb is at the focal point of the small mirror and at the center of curvature of the large mirror. in front of a mirror that has. What is the radius of curvature 44. An object is placed a magnification of of the mirror? a. b. c. d. 16.2 Refraction 45. A scuba diver training in a pool looks at his instructor, as shown in this figure. The angle between the ray in the water and the normal to the water is 25°. Access for free at openstax.org. angle is less than 90°. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. d. When the interior material has a greater
index of refraction than the surrounding material, the incident ray may approach the boundary at an angle (called the critical angle) such that the refraction angle is 90°. The refracted ray cannot leave the interior, so it is reflected back inside and total internal reflection occurs. 16.3 Lenses 47. The muscles that change the shape of the lens in the eyes have become weak, causing vision problems for a person. In particular, the muscles cannot pull hard enough on the edges of the lens to make it less convex. Part A—What condition does inability cause? Part B—Where are images focused with respect to the retina? Part C—Which type of lens corrects this person’s problem? Explain. a. Part A—This condition causes hyperopia. Part B—Images are focused between the lens and the retina. Chapter 16 • Test Prep 521 Part C—A converging lens gathers the rays slightly so they focus onto the retina. b. Part A—This condition causes myopia. Part B—Images are focused between the lens and the retina. Part C—A converging lens gathers the rays slightly so they focus onto the retina. c. Part A—This condition causes hyperopia. Part B—Images are focused between the lens and the retina. Part C—A diverging lens spreads the rays slightly so they focus onto the retina. d. Part A—This condition causes myopia. Part B—Images are focused between the lens and the retina. Part C—A diverging lens spreads the rays slightly so they focus onto the retina. 48. If the lens-to-retina distance is, what is the power of the eye when viewing an object a. b. c. d. away? 522 Chapter 16 • Test Prep Access for free at openstax.org. CHAPTER 17 Diffraction and Interference Figure 17.1 The colors reflected by this compact disc vary with angle and are not caused by pigments. Colors such as these are direct evidence of the wave character of light. (credit: Reggie Mathalone) Chapter Outline 17.1 Understanding Diffraction and Interference 17.2 Applications of Diffraction, Interference, and Coherence Examine a compact disc under white light, noting the colors observed and their locations on the disc. Using INTRODUCTION the CD, explore the spectra of a few light sources, such as a candle flame, an incandescent bulb, and fluorescent light. If you have ever looked at
the reds, blues, and greens in a sunlit soap bubble and wondered how straw-colored soapy water could produce them, you have hit upon one of the many phenomena that can only be explained by the wave character of light. That and other interesting phenomena, such as the dispersion of white light into a rainbow of colors when passed through a narrow slit, cannot be explained fully by geometric optics. In such cases, light interacts with small objects and exhibits its wave characteristics. The topic of this chapter is the branch of optics that considers the behavior of light when it exhibits wave characteristics. 17.1 Understanding Diffraction and Interference Section Learning Objectives By the end of this section, you will be able to do the following: • Explain wave behavior of light, including diffraction and interference, including the role of constructive and destructive interference in Young’s single-slit and double-slit experiments • Perform calculations involving diffraction and interference, in particular the wavelength of light using data from a two-slit interference pattern Section Key Terms diffraction Huygens’s principle monochromatic wavefront 524 Chapter 17 • Diffraction and Interference Diffraction and Interference We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation m/s is the speed of light in vacuum, fis the frequency of the electromagnetic wave in Hz (or s–1), and where is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave. In Figure 17.2, both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves. Figure 17.2 (a) The light beam emitted by a laser at the Paranal Observatory (part
of the European Southern Observatory in Chile) acts like a ray, traveling in a straight line. (credit: Yuri Beletsky, European Southern Observatory) (b) A laser beam passing through a grid of vertical slits produces an interference pattern—characteristic of a wave. (credit: Shim’on and Slava Rybka, Wikimedia Commons) That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves. Figure 17.3 shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern. Figure 17.3 Incoming waves (at the top of the picture) pass through the gaps in the rocks and create an interference pattern (in the foreground). Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f, remains the same. The speed of light in a medium is where nis its index of refraction. If you divide both sides of the equation is the wavelength in a medium, and by n, you get. Therefore,, where, Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 525 where is the wavelength in vacuum and nis the medium’s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n= 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or not, since colors are associated with frequency. 285–570 nm. Although wavelengths change while traveling from one medium to another, colors do The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts, which are the points on a wave’s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens’s principle states, “Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same
speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.” Figure 17.4 shows how Huygens’s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn later at a time, t, so that they have moved a distance. The new wavefront is a line tangent to the wavelets and is where the wave is located at time t. Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere. Figure 17.4 Huygens’s principle applied to a straight wavefront. Each point on the wavefront emits a semicircular wavelet that moves a distance. The new wavefront is a line tangent to the wavelets. What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners. If light passes through smaller openings, often called slits, you can use Huygens’s principle to show that light bends as sound does (see Figure 17.5). The bending of a wave around the edges of an opening or an obstacle is called diffraction. Diffraction is a 526 Chapter 17 • Diffraction and Interference wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in Figure 17.2 is evidence that light has the properties of a wave. Figure 17.5 Huy
gens’s principle applied to a straight wavefront striking an opening. The edges of the wavefront bend after passing through the opening, a process called diffraction. The amount of bending is more extreme for a small opening, consistent with the fact that wave characteristics are most noticeable for interactions with objects about the same size as the wavelength. Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in Figure 17.6. Figure 17.6 Ocean waves pass through an opening in a reef, resulting in a diffraction pattern. Diffraction occurs because the opening is similar in width to the wavelength of the waves. WATCH PHYSICS Single-Slit Interference This video works through the math needed to predict diffraction patterns that are caused by single-slit interference. Click to view content (https://www.openstax.org/l/28slit) Which values of mdenote the location of destructive interference in a single-slit diffraction pattern? a. whole integers, excluding zero b. whole integers c. d. real numbers excluding zero real numbers The fact that Huygens’s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light’s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773–1829) did his now-classic double-slit experiment (see Figure 17.7). Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 527 Figure 17.7 Young’s double-slit experiment. Here, light of a single wavelength passes through a pair of vertical slits and produces a diffraction pattern on the screen—numerous vertical light and dark lines that are spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen. When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 17.8 (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in Figure 17.8
(b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below. Figure 17.8 Double slits produce two sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. The waves overlap and interfere constructively (bright lines) and destructively (dark regions). You can only see the effect if the light falls onto a screen and is scattered into your eyes. (b) The double-slit interference pattern for water waves is nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. Virtual Physics Wave Interference Click to view content (https://www.openstax.org/l/28interference) This simulation demonstrates most of the wave phenomena discussed in this section. First, observe interference between two sources of electromagnetic radiation without adding slits. See how water waves, sound, and light all show interference patterns. Stay with light waves and use only one source. Create diffraction patterns with one slit and then with two. You may have to adjust slit width to see the pattern. 528 Chapter 17 • Diffraction and Interference Visually compare the slit width to the wavelength. When do you get the best-defined diffraction pattern? a. when the slit width is larger than the wavelength b. when the slit width is smaller than the wavelength c. when the slit width is comparable to the wavelength d. when the slit width is infinite Calculations Involving Diffraction and Interference The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young’s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a
whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More, then generally, if the paths taken by the two waves differ by any half-integral number of wavelengths destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths, then constructive interference occurs. Figure 17.9 shows how to determine the path-length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference,, where dis the distance between the slits,, to be To obtain constructive interference for a double slit, the path-length difference must be an integral multiple of the wavelength, or Similarly, to obtain destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength, or The number mis the orderof the interference. For example, m= 4 is fourth-order interference. Figure 17.9 The paths from each slit to a common point on the screen differ by an amount, assuming the distance to the screen is much greater than the distance between the slits (not to scale here). Figure 17.10 shows how the intensity of the bands of constructive interference decreases with increasing angle. Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 529 Figure 17.10 The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit. Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits. Figure 17.11 shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side. Figure 17.11 (a) Single-slit diffraction pattern. Monochromatic light passing through a single slit produces a central maximum and many smaller and dimmer maxima on either side. The central maximum is six times higher than shown. (b) The drawing shows the bright central maximum and dimmer and thinner maxima on either side. (c) The location of the minima
are shown in terms of and D. The analysis of single-slit diffraction is illustrated in Figure 17.12. Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference. or relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or farther than the ray from the top edge of the slit, they arrive out When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle out of phase. Thus, a ray from the center travels a distance of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray. farther to the common point on the screen, and so 530 Chapter 17 • Diffraction and Interference Figure 17.12 Equations for a single-slit diffraction pattern, where λis the wavelength of light, Dis the slit width, is the angle between a line from the slit to a minimum and a line perpendicular to the screen, Lis the distance from the slit to the screen, yis the distance from the center of the pattern to the minimum, and mis a nonzero integer indicating the order of the minimum. Below we summarize the equations needed for the calculations to follow. The speed of light in a vacuum, c, the wavelength of the light,, and its frequency, f, are related as follows. The wavelength of light in a medium,, compared to its wavelength in a vacuum,, is given by To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m, of the wavelength. 17.1 where dis the distance between the slits and is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength: For a single-slit diffraction pattern, the width of the slit, D, the distance of the first (m
= 1) destructive interference minimum, y, the distance from the slit to the screen, L, and the wavelength,, are given by Also, for single-slit diffraction, is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and mis the order of where the minimum. WORKED EXAMPLE Two-Slit Interference Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light? STRATEGY The third bright line is due to third-order constructive interference, which means that m= 3. You are given d= 0.0100 mm and = 10.95º. The wavelength can thus be found using the equation for constructive interference. Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 531 Solution The equation is. Solving for the wavelength,, gives Substituting known values yields 17.2 17.3 Discussion To three digits, 633 nm is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did that for visible wavelengths. His analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with, so spectra (measurements of intensity versus wavelength) can be obtained. WORKED EXAMPLE Single-Slit Diffraction Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0° relative to the incident direction of the light. What is the width of the slit? STRATEGY From the given information, and assuming the screen is far away from the slit, you can use the equation D. to find Solution Quantities given are = 550 nm, m= 2, and gives = 45.0°. Solving the equation for Dand substituting known values Discussion You see that the slit is narrow (it is only a few times greater than the wavelength of light). That is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects, such as this single-slit diffraction
pattern. Practice Problems 1. Monochromatic light from a laser passes through two slits separated by. The third bright line on a screen is 17.4 relative to the incident beam. What is the wavelength of the light? formed at an angle of a. b. c. d. 2. What is the width of a single slit through which 610-nm orange light passes to form a first diffraction minimum at an angle of 30.0°? a. 0.863 µm b. 0.704 µm c. 0.610 µm d. 1.22 µm 532 Chapter 17 • Diffraction and Interference Check Your Understanding 3. Which aspect of a beam of monochromatic light changes when it passes from a vacuum into water, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 4. Go outside in the sunlight and observe your shadow. It has fuzzy edges, even if you do not. Is this a diffraction effect? Explain. a. This is a diffraction effect. Your whole body acts as the origin for a new wavefront. b. This is a diffraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. c. This is a refraction effect. Your whole body acts as the origin for a new wavefront. d. This is a refraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. 5. Which aspect of monochromatic green light changes when it passes from a vacuum into diamond, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 17.2 Applications of Diffraction, Interference, and Coherence Section Learning Objectives By the end of this section, you will be able to do the following: • Explain behaviors of waves, including reflection, refraction, diffraction, interference, and coherence, and describe applications based on these behaviors • Perform calculations related to applications based on wave properties of light Section Key Terms differential interference contrast (DIC) diffraction grating iridescence laser monochromator Rayleigh criterion resolution Wave-Based Applications of Light In 1917, Albert Einstein was thinking about photons and excited atoms. He considered an atom excited by
a certain amount of energy and what would happen if that atom were hit by a photon with the same amount of energy. He suggested that the atom would emit a photon with that amount of energy, and it would be accompanied by the original photon. The exciting part is that you would have twophotons with the same energy andthey would be in phase. Those photons could go on to hit other excited atoms, and soon you would have a stream of in-phase photons. Such a light stream is said to be coherent. Some four decades later, Einstein’s idea found application in a process called, light amplification by stimulated emission of radiation. Take the first letters of all the words (except byand “of”) and write them in order. You get the word laser(see (a)), which is the name of the device that produces such a beam of light. Laser beams are directional, very intense, and narrow (only about 0.5 mm in diameter). These properties lead to a number of applications in industry and medicine. The following are just a few examples: • This chapter began with a picture of a compact disc (see ). Those audio and data-storage devices began replacing cassette tapes during the 1990s. CDs are read by interpreting variations in reflections of a laser beam from the surface. • Some barcode scanners use a laser beam. • Lasers are used in industry to cut steel and other metals. • Lasers are bounced off reflectors that astronauts left on the Moon. The time it takes for the light to make the round trip can be used to make precise calculations of the Earth-Moon distance. • Laser beams are used to produce holograms. The name hologram means entire picture(from the Greek holo-, as in Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 533 holistic), because the image is three-dimensional. A viewer can move around the image and see it from different perspectives. Holograms take advantage of the wave properties of light, as opposed to traditional photography which is based on geometric optics. A holographic image is produced by constructive and destructive interference of a split laser beam. • One of the advantages of using a laser as a surgical tool is that it is accompanied by very little bleeding. • Laser eye surgery has improved the vision of many people, without the need for corrective lenses. A laser beam is used to change the shape of the lens of the eye, thus changing its
focal length. Virtual Physics Lasers Click to view content (https://www.openstax.org/l/28lasers) This animation allows you to examine the workings of a laser. First view the picture of a real laser. Change the energy of the incoming photons, and see if you can match it to an excitation level that will produce pairs of coherent photons. Change the excitation level and try to match it to the incoming photon energy. In the animation there is only one excited atom. Is that the case for a real laser? Explain. a. No, a laser would have two excited atoms. b. No, a laser would have several million excited atoms. c. Yes, a laser would have only one excited atom. d. No, a laser would have on the order of 1023 excited atoms. An interesting thing happens if you pass light through a large number of evenly-spaced parallel slits. Such an arrangement of slits is called a diffraction grating. An interference pattern is created that is very similar to the one formed by double-slit diffraction (see and ). A diffraction grating can be manufactured by scratching glass with a sharp tool to form a number of precisely positioned parallel lines, which act like slits. Diffraction gratings work both for transmission of light, as in Figure 17.13, and for reflection of light, as on the butterfly wings or the Australian opal shown in Figure 17.14, or the CD pictured in the opening illustration of this chapter. In addition to their use as novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than do double slits. That is, their bright regions are narrower and brighter, while their dark regions are darker. Figure 17.15 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. The effect is called iridescence. Figure 17.13 A diffraction grating consists of a large number of evenly-spaced parallel slits. (a) Light passing through the grating is diffracted in a pattern similar to a double slit, with bright regions at various angles. (b) The pattern obtained for white light incident on a grating. The
central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors. 534 Chapter 17 • Diffraction and Interference Figure 17.14 (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credit: (a) Opals-On-Black.com, via Flickr (b) whologwhy, Flickr) Figure 17.15 Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the same angles, but those for the diffraction grating are narrower, and hence sharper. The maxima become narrower and the regions between become darker as the number of slits is increased. Snap Lab Diffraction Grating • A CD (compact disc) or DVD • A measuring tape • Sunlight near a white wall Instructions Procedure 1. Hold the CD in direct sunlight near the wall, and move it around until a circular rainbow pattern appears on the wall. 2. Measure the distance from the CD to the wall and the distance from the center of the circular pattern to a color in the rainbow. Use those two distances to calculate. Find.. 3. Look up the wavelength of the color you chose. That is 4. Solve 5. Compare your answer to the usual spacing between CD tracks, which is 1,600 nm (1.6 μm). for d. How do you know what number to use for m? a. Count the rainbow rings preceding the chosen color. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 535 b. Calculate mfrom the frequency of the light of the chosen color. c. Calculate mfrom the wavelength of the light of the chosen color. d. The value of mis fixed for every color. FUN IN PHYSICS CD Players Can you see the grooves on a CD or DVD (see Figure 17.16)? You may think you can because you know they are there, but they are extremely narrow—1,600 in a millimeter. Because the width of the grooves is similar to wavelengths of visible light, they form a diffraction grating. That is why you see rainbows on a CD. The colors are attractive, but they are incidental to the functions of storing and retrieving audio and other
data. Figure 17.16 For its size, this CD holds a surprising amount of information. Likewise, the CD player it is in houses a surprising number of electronic devices. The grooves are actually one continuous groove that spirals outward from the center. Data are recorded in the grooves as binary code (zeroes and ones) in small pits. Information in the pits is detected by a laser that tracks along the groove. It gets even more complicated: The speed of rotation must be varied as the laser tracks toward the circumference so that the linear speed along the groove remains constant. There is also an error correction mechanism to prevent the laser beam from getting off track. A diffraction grating is used to create the first two maxima on either side of the track. If those maxima are not the same distance from the track, an error is indicated and then corrected. The pits are reflective because they have been coated with a thin layer of aluminum. That allows the laser beam to be reflected back and directed toward a photodiode detector. The signal can then be processed and converted to the audio we hear. The longest wavelength of visible light is about 780 nm. How does that compare to the distance between CD grooves? a. The grooves are about 3 times the longest wavelength of visible light. b. The grooves are about 2 times the longest wavelength of visible light. c. The grooves are about 2 times the shortest wavelength of visible light. d. The grooves are about 3 times the shortest wavelength of visible light. LINKS TO PHYSICS Biology: DIC Microscopy If you were completely transparent, it would be hard to recognize you from your photograph. The same problem arises when using a traditional microscope to view or photograph small transparent objects such as cells and microbes. Microscopes using differential interference contrast (DIC) solve the problem by making it possible to view microscopic objects with enhanced contrast, as shown in Figure 17.17. 536 Chapter 17 • Diffraction and Interference Figure 17.17 This aquatic organism was photographed with a DIC microscope. (credit: Public Library of Science) A DIC microscope separates a polarized light source into two beams polarized at right angles to each other and coherent with each other, that is, in phase. After passing through the sample, the beams are recombined and realigned so they have the same plane of polarization. They then create an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed
through. The result is an image with contrast and shadowing that could not be observed with traditional optics. Where are diffraction gratings used? Diffraction gratings are key components of monochromators—devices that separate the various wavelengths of incoming light and allow a beam with only a specific wavelength to pass through. Monochromators are used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength of light emitted by molecules in diseased cells in a biopsy sample, or to help excite strategic molecules in the sample with a selected frequency of light. Another important use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings is available for selecting specific wavelengths for such use. Diffraction gratings are used in spectroscopes to separate a light source into its component wavelengths. When a material is heated to incandescence, it gives off wavelengths of light characteristic of the chemical makeup of the material. A pure substance will produce a spectrum that is unique, thus allowing identification of the substance. Spectroscopes are also used to measure wavelengths both shorter and longer than visible light. Such instruments have become especially useful to astronomers and chemists. Figure 17.18 shows a diagram of a spectroscope. Figure 17.18 The diagram shows the function of a diffraction grating in a spectroscope. Light diffracts as it moves through space, bending around obstacles and interfering constructively and destructively. While diffraction allows light to be used as a spectroscopic tool, it also limits the detail we can obtain in images. Figure 17.19 (a) shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. This pattern is caused by diffraction similar to that produced by a single slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large apertures, too. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 537 Figure 17.19 (a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point light sources that are close to one another produce overlapping images because of diff
raction. (c) If they are closer together, they cannot be resolved, that is, distinguished. How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 17.19 (b) shows the diffraction pattern produced by two point light sources that are close to one another. The pattern is similar to that for a single point source, and it is just barely possible to tell that there are two light sources rather than one. If they are closer together, as in Figure 17.19 (c), you cannot distinguish them, thus limiting the detail, or resolution, you can obtain. That limit is an inescapable consequence of the wave nature of light. There are many situations in which diffraction limits the resolution. The acuity of vision is limited because light passes through the pupil, the circular aperture of the eye. Be aware that the diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus light passing through a lens with a diameter of Dshows the diffraction effect and spreads, blurring the image, just as light passing through an aperture of diameter Ddoes. Diffraction limits the resolution of any system having a lens or mirror. Telescopes are also limited by diffraction, because of the finite diameter, D, of their primary mirror. Why are diffraction gratings used in spectroscopes rather than just two slits? a. The bands produced by diffraction gratings are dimmer but sharper than the bands produced by two slits. b. The bands produced by diffraction gratings are brighter, though less sharp, than the bands produced by two slits. c. The bands produced by diffraction gratings are brighter and sharper than the bands produced by two slits. d. The bands produced by diffraction gratings are dimmer and less sharp, but more widely dispersed, than the bands produced by two slits. Calculations Involving Diffraction Gratings and Resolution Early in the chapter, it was mentioned that when light passes from one medium to another, its speed and wavelength change, but its frequency remains constant. The equation, is related to the wavelength in a vacuum, shows how to the wavelength in a given medium,, and the refractive index, n, of the medium. The equation is useful for calculating the change in wavelength of a monochromatic laser beam in various media. The analysis of a diffraction grating is very similar to that for a
double slit. As you know from the discussion of double slits in Young’s double-slit experiment, light is diffracted by, and spreads out after passing through, each slit. Rays travel at an angle relative to the incident direction. Each ray travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled. Each ray travels a distance that differs by equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is from that of its neighbor, where dis the distance between slits. If where dis the distance between slits in the grating, is the wavelength of the light, and mis the order of the maximum. Note that this is exactly the same equation as for two slits separated by d. However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles. 538 Chapter 17 • Diffraction and Interference WATCH PHYSICS Diffraction Grating This video (https://www.openstax.org/l/28diffraction) explains the geometry behind the diffraction pattern produced by a diffraction grating. Click to view content (https://www.openstax.org/l/28diffraction) The equation that gives the points of constructive interference produced by a diffraction grating is equation look familiar? a. b. c. d. It is the same as the equation for destructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a single-slit diffraction pattern. It is the same as the equation for destructive interference for a single-slit diffraction pattern.. Why does that Just what is the resolution limit of an aperture or lens? To answer that question, consider the diffraction pattern for a circular aperture, which, similar to the diffraction pattern of light passing through a slit, has a central maximum that is wider and brighter than the maxima surrounding it (see Figure 17.19 (a)). It can be shown that, for a circular aperture of diameter D, the first minimum in the diffraction pattern occurs at wavelength of
light, which is the case for most optical instruments. The accepted criterion for determining the diffraction limit to resolution based on diffraction was developed by Lord Rayleigh in the 19th century. The Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 17.20 (b). The first minimum is at an angle of, so that two point objects are just resolvable if they are separated by the angle, provided that the aperture is large compared with the is the wavelength of the light (or other electromagnetic radiation) and Dis the diameter of the aperture, lens, mirror, where etc., with which the two objects are observed. In the expression above, has units of radians. Figure 17.20 (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for their being just resolvable. The central maximum of one pattern lies on the first minimum of the other. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 539 Snap Lab Resolution • A sheet of white paper • A black pen or pencil • A measuring tape Instructions Procedure 1. Draw two lines several mm apart on a white sheet of paper. 2. Move away from the sheet as it is held upright, and measure the distance at which you can just distinguish (resolve) the lines as separate. 3. Use to calculate Dthe diameter of your pupil. Use the distance between the lines and the maximum distance at which they were resolved to calculate. Use the average wavelength for visible light as the value for. 4. Compare your answer to the average pupil diameter of 3 mm. Describe resolution in terms of minima and maxima of diffraction patterns. a. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. b. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. c. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the
second minimum of the pattern for the other line. d. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the second maximum of the pattern for the other line. WORKED EXAMPLE Change of Wavelength A monochromatic laser beam of green light with a wavelength of 550 nm passes from air to water. The refractive index of water is 1.33. What will be the wavelength of the light after it enters the water? STRATEGY You can assume that the refractive index of air is the same as that of light in a vacuum because they are so close. You then have all the information you need to solve for. Solution 17.5 Discussion The refractive index of air is 1.0003, so the approximation holds for three significant figures. You would not see the light change color, however. Color is determined by frequency, not wavelength. WORKED EXAMPLE Diffraction Grating A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum form for green light with a wavelength of 520 nm? STRATEGY You are given enough information to calculate d, and you are given the values of and m. You will have to find the arcsin of a 540 Chapter 17 • Diffraction and Interference number to find. Solution First find d. Rearrange the equation for constructive interference conditions for a diffraction grating, and substitute the known values. 17.6 Discussion This angle seems reasonable for the first maximum. Recall that the meaning of sin‒1 (or arcsin) is the angle with a sine that is (the unknown). Remember that the value of will not be greater than 1 for any value of. WORKED EXAMPLE Resolution What is the minimum angular spread of a 633-nm-wavelength He-Ne laser beam that is originally 1.00 mm in diameter? STRATEGY The diameter of the beam is the same as if it were coming through an aperture of that size, so D= 1.00 mm. You are given, and you must solve for. Solution 17.7 Discussion The conversion factor for radians to degrees is 1.000 radian = 57.3°. The spread is very small and would not be noticeable over short distances. The angle represents the angular separation of the central maximum and the first minimum. Practice Problems 6. A beam of yellow light has a wavelength of 600 nm in a vacuum and a wavelength of 397 nm in Plex
iglas. What is the refractive index of Plexiglas? a. 1.51 b. 2.61 3.02 c. 3.77 d. 7. What is the angle between two just-resolved points of light for a 3.00 mm diameter pupil, assuming an average wavelength of 550 nm? a. 224 rad 183 rad b. 1.83 × 10–4 rad c. d. 2.24 × 10–4 rad Check Your Understanding 8. How is an interference pattern formed by a diffraction grating different from the pattern formed by a double slit? Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 541 a. The pattern is colorful. b. The pattern is faded. c. The pattern is sharper. d. The pattern is curved. 9. A beam of light always spreads out. Why can a beam not be produced with parallel rays to prevent spreading? a. Light is always polarized. b. Light is always reflected. c. Light is always refracted. d. Light is always diffracted. 10. Compare interference patterns formed by a double slit and by a diffraction grating in terms of brightness and narrowness of bands. a. The pattern formed has broader and brighter bands. b. The pattern formed has broader and duller bands. c. The pattern formed has narrower and duller bands. d. The pattern formed has narrower and brighter bands. 11. Describe the slits in a diffraction grating in terms of number and spacing, as compared to a two-slit diffraction setup. a. The slits in a diffraction grating are broader, with space between them that is greater than the separation of the two slits in two-slit diffraction. b. The slits in a diffraction grating are broader, with space between them that is the same as the separation of the two slits in two-slit diffraction. c. The slits in a diffraction grating are narrower, with space between them that is the same as the separation of the two slits in two-slit diffraction. d. The slits in a diffraction grating are narrower, with space between them that is greater than the separation of the two slits in two-slit diffraction. 542 Chapter 17 • Key Terms KEY TERMS differential interference contrast (DIC) separating a tangent to all of the
wavelets. polarized light source into two beams polarized at right angles to each other and coherent with each other then, after passing through the sample, recombining and realigning the beams so they have the same plane of polarization, and then creating an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through; the result is an image with contrast and shadowing that could not be observed with traditional optics diffraction bending of a wave around the edges of an opening or an obstacle diffraction grating many of evenly spaced slits having dimensions such that they produce an interference pattern Huygens’s principle Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself; the new wavefront is a line SECTION SUMMARY 17.1 Understanding Diffraction and Interference iridescence the effect that occurs when tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives feathers colors not solely due to their pigmentation laser acronym for a device that produces light amplification by stimulated emission of radiation monochromatic one color monochromator device that separates the various wavelengths of incoming light and allows a beam with only a specific wavelength to pass through Rayleigh criterion two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other resolution degree to which two images can be distinguished from one another, which is limited by diffraction wavefront points on a wave surface that all share an identical, constant phase 17.2 Applications of Diffraction, Interference, and Coherence • The wavelength of light varies with the refractive index • The focused, coherent radiation emitted by lasers has of the medium. many uses in medicine and industry. • Slits produce a diffraction pattern if their width and • Characteristics of diffraction patterns produced with separation are similar to the wavelength of light passing through them. Interference bands of a single-slit diffraction pattern can be predicted. Interference bands of a double-slit diffraction pattern can be predicted. • • KEY EQUATIONS 17.1 Understanding Diffraction and Interference speed of light, frequency, and wavelength change of wavelength with index of refraction two-slit constructive interference, for m = 0, 1, −1, 2, −2, … diffraction gratings can be determined. • Diffraction gratings have been
incorporated in many instruments, including microscopes and spectrometers. • Resolution has a limit that can be predicted. two-slit destructive interference, for m= 0, 1, −1, 2, −2, … one-slit, first-order destructive interference; wavelength related to dimensions one-slit destructive interference Access for free at openstax.org. Chapter 17 • Chapter Review 543 17.2 Applications of Diffraction, Interference, and Coherence wavelength change with change in medium diffraction grating constructive interference resolution CHAPTER REVIEW Concept Items 17.1 Understanding Diffraction and Interference 1. Which behavior of light is indicated by an interference pattern? a. ray behavior b. particle behavior c. d. wave behavior corpuscular behavior 2. Which behavior of light is indicated by diffraction? a. wave behavior b. particle behavior ray behavior c. corpuscular behavior d. 17.2 Applications of Diffraction, Interference, and Coherence 3. There is a principle related to resolution that is expressed by this equation. 17.8 What is that principle stated in full? Critical Thinking Items 17.1 Understanding Diffraction and Interference 6. Describe a situation in which bodies of water and a line of rocks could create a diffraction pattern similar to light passing through double slits. Include the arrangement of the rocks, the positions of the bodies of water, and the location of the diffraction pattern. Note the dimensions that are necessary for the production of the pattern. a. When waves from a small body of water pass through two widely separated openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The width of each opening is larger than the size of the wavelength of the waves. b. When waves from a large body of water pass 4. A principle related to resolution states, “Two images are just resolved when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.” Write the equation that expresses that principle. a. b. c. d. 5. Which statement completes this resolution? Two images are just resolved when — a. The center of the diffraction pattern of one image is directly over the central maximum of the diffraction pattern of the other. b. The center of the diffraction pattern of one image is directly over the central minimum of the diffraction pattern of the other c. The
center of the diffraction pattern of one image is directly over the first minimum of the diffraction pattern of the other d. The center of the diffraction pattern of one is directly over the first maximum of the diffraction pattern of the other through two narrow openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and separation of the openings are similar to the size of the wavelength of the waves. c. When waves from a small body of water pass through two wide openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The separation between the openings is similar to the size of the wavelength of the waves. d. When waves from a large body of water pass through two wide openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and 544 Chapter 17 • Chapter Review separation of the openings are larger than the size of the wavelength of the waves. 17.2 Applications of Diffraction, Interference, and Coherence 7. For what type of electromagnetic radiation would a grating with spacing greater than 800 nm be useful as a spectroscopic tool? a. It can be used to analyze spectra only in the infrared portion of the spectrum. It can be used to analyze spectra in the entire visible portion of the electromagnetic spectrum. b. Problems 17.1 Understanding Diffraction and Interference 9. What is the distance between two slits that produce a diffraction pattern with the first minimum at an angle of 45.0° when 410-nm violet light passes through the slits? a. 2,030 nm b. 1,450 nm c. 410 nm d. 290 nm 10. A breakwater at the entrance to a harbor consists of a rock barrier with a 50.0 − m -wide opening. Ocean waves with a 20.0-m wavelength approach the opening straight on. At what angle to the incident direction are the boats inside the harbor most protected against wave action? 11.5° a. 7.46° b. c. 5.74° d. 23.6° Performance Task 17.2 Applications of Diffraction, Interference, and Coherence 13. In this performance task you will create one- and two- slit diffraction and observe the
interference patterns that result. • A utility knife (a knife with a razor blade-like cutting edge) • Aluminum foil • A straight edge • A strong, small light source or a laser pointer • A tape measure • A white wall Access for free at openstax.org. c. d. It can only be used to analyze spectra in the short‐wavelength visible. It can only be used to analyze spectra in the short‐wavelength visible and ultraviolet. 8. A beam of green light has a wavelength of in a in Plexiglas. What vacuum and a wavelength of is the refractive index of Plexiglas? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 11. A 500-nm beam of light passing through a diffraction grating creates its second band of constructive interference at an angle of 1.50°. How far apart are the slits in the grating? 38,200 nm a. b. 19,100 nm c. 667 nm 333 nm d. 12. The range of the visible-light spectrum is 380 nm to 780 nm. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? a. 26,300 lines/cm b. 13,200 lines/cm c. 6,410 lines/cm d. 12,820 lines/cm Procedure 1. Cut a piece of aluminum foil about 15 cm × 15 cm. 2. Use the utility knife and the straight edge to cut a straight slit several cm long in the center of the foil square. 3. With the room darkened, one partner shines the light through the slit and toward the wall. The other partner observes the pattern on the wall. The partner with the light changes the distance from the foil to the wall and the distance from the light to the foil. 4. When the sharpest, brightest pattern possible is obtained, the partner who is not holding the foil and light makes measurements. 5. Measure the perpendicular (shortest) distance from the slit to the wall, the distance from the center of the pattern to several of the dark bands, and the distance from the slit to the same dark bands. 6. Carefully make a second slit parallel to the first slit and 1 mm or less away. 7. Repeat steps 2 through 5, only this time measure the distances to bright bands. NOTE—In your calculations, use 580 nm for
if you used white light. If you used a colored laser pointer, look up the wavelength of the color. You from its tangent may find it easier to calculate TEST PREP Multiple Choice 17.1 Understanding Diffraction and Interference 14. Which remains unchanged when a monochromatic beam of light passes from air into water? a. b. c. d. the speed of the light the direction of the beam the frequency of the light the wavelength of the light 15. Two slits are separated by a distance of 3500 nm. If light with a wavelength of 500 nm passes through the slits and produces an interference pattern, the m = ________ order minimum appears at an angle of 30.0°. a. 0 b. 1 c. 2 3 d. 16. In the sunlight, the shadow of a building has fuzzy edges even if the building does not. Is this a refraction effect? Explain. a. Yes, this is a refraction effect, where every point on the building acts as the origin for a new wavefront. b. Yes, this is a refraction effect, where the whole building acts as the origin for a new wavefront. c. No, this is a diffraction effect, where every point on the edge of the building’s shadow acts as the origin for a new wavefront. d. No, this is a diffraction effect, where the whole building acts as the origin for a new wavefront. 17.2 Applications of Diffraction, Interference, and Coherence 17. Two images are just resolved when the center of the diffraction pattern of one is directly over ________ of the diffraction pattern of the other. a. the center Chapter 17 • Test Prep 545 rather than from its sine. a. Which experiment gave the most distinct pattern—one or two slits? b. What was the width of the single slit? Compare the calculated distance with the measured distance. c. What was the distance between the two slits? Compare the calculated distance with the measured distance. b. c. d. the first minimum the first maximum the last maximum 18. Two point sources of light are just resolvable as. What is the they pass through a small hole. The angle to the first minimum of one source is diameter of the hole? a. b. c. d. 19. Will a beam of light shining through a 1-mm hole behave any differently than a beam of light that is 1 mm wide as it leaves
its source? Explain.? a. Yes, the beam passing through the hole will spread out as it travels, because it is diffracted by the edges of the hole, whereas the 1 -mm beam, which encounters no diffracting obstacle, will not spread out. b. Yes, the beam passing through the hole will be made more parallelby passing through the hole, and so will not spread out as it travels, whereas the unaltered wavefronts of the 1-mm beam will cause the beam to spread out as it travels. c. No, both beams will remain the same width as they travel, and they will not spread out. d. No, both beams will spread out as they travel. 20. A laser pointer emits a coherent beam of parallel light rays. Does the light from such a source spread out at all? Explain. a. Yes, every point on a wavefront is not a source of wavelets, which prevent the spreading of light waves. b. No, every point on a wavefront is not a source of wavelets, so that the beam behaves as a bundles of rays that travel in their initial direction. c. No, every point on a wavefront is a source of 546 Chapter 17 • Test Prep wavelets, which keep the beam from spreading. d. Yes, every point on a wavefront is a source of wavelets, which cause the beam to spread out steadily as it moves forward. Short Answer 17.1 Understanding Diffraction and Interference 21. Light passing through double slits creates a diffraction pattern. How would the spacing of the bands in the pattern change if the slits were closer together? a. The bands would be closer together. b. The bands would spread farther apart. c. The bands would remain stationary. d. The bands would fade and eventually disappear. 22. A beam of light passes through a single slit to create a diffraction pattern. How will the spacing of the bands in the pattern change if the width of the slit is increased? a. The width of the spaces between the bands will remain the same. spaced parallel lines that produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. b. A diffraction grating is a large collection of randomly spaced parallel lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. c. A diffraction grating is a large collection of randomly spaced intersecting lines that
produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. d. A diffraction grating is a large collection of evenly spaced intersecting lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. b. The width of the spaces between the bands will 26. Suppose pure-wavelength light falls on a diffraction increase. c. The width of the spaces between the bands will decrease. d. The width of the spaces between the bands will first decrease and then increase. grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? a. The bands will spread farther from the central maximum. b. The bands will come closer to the central 23. What is the wavelength of light falling on double slits maximum. if the third-order maximum is at c. The bands will not spread farther from the first maximum. d. The bands will come closer to the first maximum. 27. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 473 nm blue light at an angle of 25.0°? a. b. c. 851 lines/cm d. 8,934 lines/cm 529,000 lines/cm 50,000 lines/cm 28. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of 60.0°? a. 2.28 × 104 nm 3.29 × 102 nm b. c. 2.53 × 101 nm 1.76 × 103 nm d.? separated by an angle of a. b. c. d. 24. What is the longest wavelength of light passing through a single slit of width 1.20 μm for which there is a firstorder minimum? 1.04 µm a. b. 0.849 µm c. 0.600 µm d. 2.40 µm 17.2 Applications of Diffraction, Interference, and Coherence 25. Describe a diffraction grating and the interference pattern it produces. a. A diffraction grating is a large collection of evenly Access for free at openstax.org. Chapter 17 • Test Prep 547 grating. a. All three interference pattern produce identical bands. b. A double slit produces the sharpest and most distinct bands
. c. A single slit produces the sharpest and most distinct bands. d. The diffraction grating produces the sharpest and most distinct bands. 32. An electric current through hydrogen gas produces several distinct wavelengths of visible light. The light is projected onto a diffraction grating having per centimeter. What are the wavelengths of the hydrogen spectrum if the light forms first-order maxima at angles of a., and?,, lines b. c. d. Extended Response 17.1 Understanding Diffraction and Interference 29. Suppose you use a double slit to perform Young’s double-slit experiment in air, and then repeat the experiment with the same double slit in water. Does the color of the light change? Do the angles to the same parts of the interference pattern get larger or smaller? Explain. a. No, the color is determined by frequency. The magnitude of the angle decreases. b. No, the color is determined by wavelength. The magnitude of the angle decreases. c. Yes, the color is determined by frequency. The magnitude of the angle increases. d. Yes, the color is determined by wavelength. The magnitude of the angle increases. 30. A double slit is located at a distance xfrom a screen, with the distance along the screen from the center given by y. When the distance dbetween the slits is relatively large, there will be numerous bright bands. For small angles (where sinθ= θ, with θin radians), what is the distance between fringes? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 31. Compare the interference patterns of single-slit diffraction, double-slit diffraction, and a diffraction 548 Chapter 17 • Test Prep Access for free at openstax.org. CHAPTER 18 Static Electricity Figure 18.1 This child’s hair contains an imbalance of electrical charge (commonly called static electricity), which causes it to stand on end. The sliding motion stripped electrons away from the child’s body, leaving him with an excess of positive charges, which repel each other along each strand of hair. (credit: Ken Bosma, Wikimedia Commons) Chapter Outline 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 18.2 Coulomb's law 18.3 Electric Field 18.4 Electric Potential 18.5 Capacitors and Dielectrics You may have been introduced to static electricity like
the child sliding down the slide in the opening INTRODUCTION photograph (Figure 18.1). The zapthat he is likely to receive if he touches a playmate or parent tends to bring home the lesson. But static electricity is more than just fun and games—it is put to use in many industries. The forces between electrically charged particles are used in technologies such as printers, pollution filters, and spray guns used for painting cars and trucks. Static electricityis the study of phenomena that involve an imbalance of electrical charge. Although creating this imbalance typically requires moving charge around, once the imbalance is created, it often remains static for a long time. The study of charge in motion is called electromagnetismand will be covered in a later chapter. What is electrical charge, how is it associated 550 Chapter 18 • Static Electricity with objects, and what forces does it create? These are just some of the questions that this chapter addresses. 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge Section Learning Objectives By the end of this section, you will be able to do the following: • Describe positive and negative electric charges • Use conservation of charge to calculate quantities of charge transferred between objects • Characterize materials as conductors or insulators based on their electrical properties • Describe electric polarization and charging by induction Section Key Terms conduction conductor electron induction insulator law of conservation of charge polarization proton Electric Charge You may know someone who has an electricpersonality, which usually means that other people are attracted to this person. This saying is based on electric charge, which is a property of matter that causes objects to attract or repel each other. Electric charge comes in two varieties, which we call positiveand negative.Like charges repel each other, and unlike charges attract each other. Thus, two positive charges repel each other, as do two negative charges. A positive charge and a negative charge attract each other. How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always result in a net charge of one type on one material and a net charge of the opposite type on the other material. By convention, we call one type of charge positive and the other type negative. For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Because the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with
silk in this manner will repel one another, because each rod has positive charge on it. Similarly, two silk cloths rubbed in this manner will repel each other, because both cloths have negative charge. Figure 18.2 shows how these simple materials can be used to explore the nature of the force between charges. Figure 18.2 A glass rod becomes positively charged when rubbed with silk, whereas the silk becomes negatively charged. (a) The glass rod is attracted to the silk, because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel. It took scientists a long time to discover what lay behind these two types of charges. The word electricitself comes from the Greek word elektronfor amber, because the ancient Greeks noticed that amber, when rubbed by fur, attracts dry straw. Almost 2,000 years later, the English physicist William Gilbert proposed a model that explained the effect of electric charge as being due to a mysterious electrical fluid that would pass from one object to another. This model was debated for several hundred years, but it was finally put to rest in 1897 by the work of the English physicist J. J. Thomson and French physicist Jean Perrin. Along with many others, Thomson and Perrin were studying the mysterious cathode raysthat were known at the time to consist of particles smaller than the smallest atom. Perrin showed that cathode rays actually carried negative electrical charge. Later, Thomson’s work led him to declare, “I can see no escape from the conclusion that [cathode rays] are charges of negative Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 551 electricity carried by particles of matter.” It took several years of further experiments to confirm Thomson’s interpretation of the experiments, but science had in fact discovered the particle that carries the fundamental unit of negative electrical charge. We now know this particle as the electron. Atoms, however, were known to be electrically neutral, which means that they carry the same amount of positive and negative charge, so their net charge is zero. Because electrons are negative, some other part of the atom must contain positive charge. Thomson put forth what is called the plum pudding model, in which he described atoms as being made of thousands of electrons swimming around in a nebulous mass of positive charge, as shown by the left-side image of Figure 18.3
. His student, Ernest Rutherford, originally believed that this model was correct and used it (along with other models) to try to understand the results of his experiments bombarding gold foils with alphaparticles (i.e., helium atoms stripped of their electrons). The results, however, did not confirm Thomson’s model but rather destroyed it! Rutherford found that most of the space occupied by the gold atoms was actually empty and that almost all of the matter of each atom was concentrated into a tiny, extremely dense nucleus, as shown by the right-side image of Figure 18.3. The atomic nucleus was later found to contain particles called protons, each of which carries a unit of positive electric charge.1 Figure 18.3 The left drawing shows Thompson’s plum-pudding model, in which the electrons swim around in a nebulous mass of positive charge. The right drawing shows Rutherford’s model, in which the electrons orbit around a tiny, massive nucleus. Note that the size of the nucleus is vastly exaggerated in this drawing. Were it drawn to scale with respect to the size of the electron orbits, the nucleus would not be visible to the naked eye in this drawing. Also, as far as science can currently detect, electrons are point particles, which means that they have no size at all! Protons and electrons are thus the fundamental particles that carry electric charge. Each proton carries one unit of positive charge, and each electron carries one unit of negative charge. To the best precision that modern technology can provide, the charge carried by a proton is exactlythe opposite of that carried by an electron. The SI unit for electric charge is the coulomb (abbreviated as “C”), which is named after the French physicist Charles Augustin de Coulomb, who studied the force between charged objects. The proton carries protons required to make +1.00 C is and the electron carries. The number nof The same number of electrons is required to make −1.00 C of electric charge. The fundamental unit of charge is often represented as e. Thus, the charge on a proton is e, and the charge on an electron is −e. Mathematically, 18.1 LINKS TO PHYSICS Measuring the Fundamental Electric Charge The American physicist Robert Millikan (1868–1953) and his student Harvey Fletcher (1884–1981) were the first to make a relatively accurate measurement of the fundamental unit of charge on the electron.
They designed what is now a classic 1Protons were later found to contain sub particles called quarks, which have fractional electric charge. But that is another story that we leave for subsequent physics courses. 552 Chapter 18 • Static Electricity experiment performed by students. The Millikan oil-drop experiment is shown in Figure 18.4. The experiment involves some concepts that will be introduced later, but the basic idea is that a fine oil mist is sprayed between two plates that can be charged with a known amount of opposite charge. Some oil drops accumulate some excess negative charge when being sprayed and are attracted to the positive charge of the upper plate and repelled by the negative charge on the lower plate. By tuning the charge on these plates until the weight of the oil drop is balanced by the electric forces, the net charge on the oil drop can be determined quite precisely. Figure 18.4 The oil-drop experiment involved spraying a fine mist of oil between two metal plates charged with opposite charges. By knowing the mass of the oil droplets and adjusting the electric charge on the plates, the charge on the oil drops can be determined with precision. Millikan and Fletcher found that the drops would accumulate charge in discrete units of about within 1 percent of the modern value of times greater than the possible error Millikan reported for his results! which is Although this difference may seem quite small, it is actually five Because the charge on the electron is a fundamental constant of nature, determining its precise value is very important for all of science. This created pressure on Millikan and others after him that reveals some equally important aspects of human nature. First, Millikan took sole credit for the experiment and was awarded the 1923 Nobel Prize in physics for this work, although his student Harvey Fletcher apparently contributed in significant ways to the work. Just before his death in 1981, Fletcher divulged that Millikan coerced him to give Millikan sole credit for the work, in exchange for which Millikan promoted Fletcher’s career at Bell Labs. Another great scientist, Richard Feynman, points out that many scientists who measured the fundamental charge after Millikan were reluctant to report values that differed much from Millikan’s value. History shows that later measurements slowly crept up from Millikan’s value until settling on the modern value. Why did they not immediately find the error and correct the value, asks Feynman. Apparently, having found a value higher than the much-respected value found by Millikan, scientists would look for possible mistakes that might lower their value to
make it agree better with Millikan’s value. This reveals the important psychological weight carried by preconceived notions and shows how hard it is to refute them. Scientists, however devoted to logic and data they may be, are apparently just as vulnerable to this aspect of human nature as everyone else. The lesson here is that, although it is good to be skeptical of new results, you should not discount them just because they do not agree with conventional wisdom. If your reasoning is sound and your data are reliable, the conclusion demanded by the data must be seriously considered, even if that conclusion disagrees with the commonly accepted truth. GRASP CHECK Suppose that Millikan observed an oil drop carrying three fundamental units of charge. What would be the net charge on this oil drop? a. −4.81 × 10−19 C b. −1.602 × 10−19 C 1.602 × 10−19 C c. d. 4.81 × 10−19 C Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 553 Snap Lab Like and Unlike Charges This activity investigates the repulsion and attraction caused by static electrical charge. • Adhesive tape • Nonconducting surface, such as a plastic table or chair Instructions Procedure for Part (a) 1. Prepare two pieces of tape about 4 cm long. To make a handle, double over about 0.5 cm at one end so that the sticky side sticks together. 2. Attach the pieces of tape side by side onto a nonmetallic surface, such as a tabletop or the seat of a chair, as shown in Figure 18.5(a). 3. Peel off both pieces of tape and hang them downward, holding them by the handles, as shown in Figure 18.5(b). If the tape bends upward and sticks to your hand, try using a shorter piece of tape, or simply shake the tape so that it no longer sticks to your hand. 4. Now slowly bring the two pieces of tape together, as shown in Figure 18.5(c). What happens? Figure 18.5 Procedure for Part (b) 5. Stick one piece of tape on the nonmetallic surface, and stick the second piece of tape on top of the first piece, as shown in Figure 18.6(a). 6. Slowly peel off the two pieces by pulling on the handle of the bottom piece. 7. Gently stroke your finger along the top of the second
piece of tape (i.e., the nonsticky side), as shown in Figure 18.6(b). 8. Peel the two pieces of tape apart by pulling on their handles, as shown in Figure 18.6(c). 9. Slowly bring the two pieces of tape together. What happens? Figure 18.6 GRASP CHECK In step 4, why did the two pieces of tape repel each other? In step 9, why did they attract each other? a. Like charges attract, while unlike charges repel each other. b. Like charges repel, while unlike charges attract each other. c. Tapes having positive charge repel, while tapes having negative charge attract each other. d. Tapes having negative charge repel, while tapes having positive charge attract each other. 554 Chapter 18 • Static Electricity Conservation of Charge Because the fundamental positive and negative units of charge are carried on protons and electrons, we would expect that the total charge cannot change in any system that we define. In other words, although we might be able to move charge around, we cannot create or destroy it. This should be true provided that we do not create or destroy protons or electrons in our system. In the twentieth century, however, scientists learned how to create and destroy electrons and protons, but they found that charge is still conserved. Many experiments and solid theoretical arguments have elevated this idea to the status of a law. The law of conservation of charge says that electrical charge cannot be created or destroyed. The law of conservation of charge is very useful. It tells us that the net charge in a system is the same before and after any interaction within the system. Of course, we must ensure that no external charge enters the system during the interaction and that no internal charge leaves the system. Mathematically, conservation of charge can be expressed as 18.2 where is the net charge of the system before the interaction, and is the net charge after the interaction. WORKED EXAMPLE What is the missing charge? Figure 18.7 shows two spheres that initially have +4 C and +8 C of charge. After an interaction (which could simply be that they touch each other), the blue sphere has +10 C of charge, and the red sphere has an unknown quantity of charge. Use the law of conservation of charge to find the final charge on the red sphere. Strategy The net initial charge of the system is. The net final charge of the system is is the final charge on the red sphere
. Conservation of charge tells us that we can solve for., where, so 18.3 Solution Equating and and solving for gives The red sphere has +2 C of charge. Figure 18.7 Two spheres, one blue and one red, initially have +4 C and +8 C of charge, respectively. After the two spheres interact, the blue sphere has a charge of +10 C. The law of conservation of charge allows us to find the final charge on the red sphere. Discussion Like all conservation laws, conservation of charge is an accounting scheme that helps us keep track of electric charge. Practice Problems 1. Which equation describes conservation of charge? a. qinitial = qfinal = constant b. qinitial = qfinal = 0 c. qinitial − qfinal = 0 Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 555 d. qinitial/qfinal = constant 2. An isolated system contains two objects with charges and. If object 1 loses half of its charge, what is the final charge on object 2? a. b. c. d. Conductors and Insulators Materials can be classified depending on whether they allow charge to move. If charge can easily move through a material, such as metals, then these materials are called conductors. This means that charge can be conducted (i.e., move) through the material rather easily. If charge cannot move through a material, such as rubber, then this material is called an insulator. Most materials are insulators. Their atoms and molecules hold on more tightly to their electrons, so it is difficult for electrons to move between atoms. However, it is not impossible. With enough energy, it is possible to force electrons to move through an insulator. However, the insulator is often physically destroyed in the process. In metals, the outer electrons are loosely bound to their atoms, so not much energy is required to make electrons move through metal. Such metals as copper, silver, and aluminum are good conductors. Insulating materials include plastics, glass, ceramics, and wood. The conductivity of some materials is intermediate between conductors and insulators. These are called semiconductors. They can be made conductive under the right conditions, which can involve temperature, the purity of the material, and the force applied to push electrons through them. Because we can control whether semiconductors are conductors or insulators, these materials are used extensively in computer chips.
The most commonly used semiconductor is silicon. Figure 18.8 shows various materials arranged according to their ability to conduct electrons. Figure 18.8 Materials can be arranged according to their ability to conduct electric charge. The slashes on the arrow mean that there is a very large gap in conducting ability between conductors, semiconductors, and insulators, but the drawing is compressed to fit on the page. The numbers below the materials give their resistivityin Ω•m (which you will learn about below). The resistivity is a measure of how hard it is to make charge move through a given material. What happens if an excess negative charge is placed on a conducting object? Because like charges repel each other, they will push against each other until they are as far apart as they can get. Because the charge can move in a conductor, it moves to the outer surfaces of the object. Figure 18.9(a) shows schematically how an excess negative charge spreads itself evenly over the outer surface of a metal sphere. What happens if the same is done with an insulating object? The electrons still repel each other, but they are not able to move, because the material is an insulator. Thus, the excess charge stays put and does not distribute itself over the object. Figure 18.9(b) shows this situation. 556 Chapter 18 • Static Electricity Figure 18.9 (a) A conducting sphere with excess negative charge (i.e., electrons). The electrons repel each other and spread out to cover the outer surface of the sphere. (b) An insulating sphere with excess negative charge. The electrons cannot move, so they remain in their original positions. Transfer and Separation of Charge Most objects we deal with are electrically neutral, which means that they have the same amount of positive and negative charge. However, transferring negative charge from one object to another is fairly easy to do. When negative charge is transferred from one object to another, an excess of positive charge is left behind. How do we know that the negative charge is the mobile charge? The positive charge is carried by the proton, which is stuck firmly in the nucleus of atoms, and the atoms are stuck in place in solid materials. Electrons, which carry the negative charge, are much easier to remove from their atoms or molecules and can therefore be transferred more easily. Electric charge can be transferred in several manners. One of the simplest ways to transfer charge is charging by contact, in which the surfaces of two objects made
of different materials are placed in close contact. If one of the materials holds electrons more tightly than the other, then it takes some electrons with it when the materials are separated. Rubbing two surfaces together increases the transfer of electrons, because it creates a closer contact between the materials. It also serves to present freshmaterial with a full supply of electrons to the other material. Thus, when you walk across a carpet on a dry day, your shoes rub against the carpet, and some electrons are removed from the carpet by your shoes. The result is that you have an excess of negative charge on your shoes. When you then touch a doorknob, some of your excess of electrons transfer to the neutral doorknob, creating a small spark. Touching the doorknob with your hand demonstrates a second way to transfer electric charge, which is charging by conduction. This transfer happens because like charges repel, and so the excess electrons that you picked up from the carpet want to be as far away from each other as possible. Some of them move to the doorknob, where they will distribute themselves over the outer surface of the metal. Another example of charging by conduction is shown in the top row of Figure 18.10. A metal sphere with 100 excess electrons touches a metal sphere with 50 excess electrons, so 25 electrons from the first sphere transfer to the second sphere. Each sphere finishes with 75 excess electrons. The same reasoning applies to the transfer of positive charge. However, because positive charge essentially cannot move in solids, it is transferred by moving negative charge in the opposite direction. For example, consider the bottom row of Figure 18.10. The first metal sphere has 100 excess protons and touches a metal sphere with 50 excess protons, so the second sphere transfers 25 electrons to the first sphere. These 25 extra electrons will electrically cancel 25 protons so that the first metal sphere is left with 75 excess protons. This is shown in the bottom row of Figure 18.10. The second metal sphere lost 25 electrons so it has 25 more excess protons, for a total of 75 excess protons. The end result is the same if we consider that the first ball transferred a net positive charge equal to that of 25 protons to the first ball. Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 557 Figure 18.10 In the top row, a metal sphere with 100 excess electrons transfers 25 electrons to a metal sphere with an
excess of 50 electrons. After the transfer, both spheres have 75 excess electrons. In the bottom row, a metal sphere with 100 excess protons receives 25 electrons from a ball with 50 excess protons. After the transfer, both spheres have 75 excess protons. In this discussion, you may wonder how the excess electrons originally got from your shoes to your hand to create the spark when you touched the doorknob. The answer is that noelectrons actually traveled from your shoes to your hands. Instead, because like charges repel each other, the excess electrons on your shoe simply pushed away some of the electrons in your feet. The electrons thus dislodged from your feet moved up into your leg and in turn pushed away some electrons in your leg. This process continued through your whole body until a distribution of excess electrons covered the extremities of your body. Thus your head, your hands, the tip of your nose, and so forth all received their doses of excess electrons that had been pushed out of their normal positions. All this was the result of electrons being pushed out of your feet by the excess electrons on your shoes. This type of charge separation is called polarization. As soon as the excess electrons leave your shoes (by rubbing off onto the floor or being carried away in humid air), the distribution of electrons in your body returns to normal. Every part of your body is again electrically neutral (i.e., zero excess charge). The phenomenon of polarization is seen in. The child has accumulated excess positive charge by sliding on the slide. This excess charge repels itself and so becomes distributed over the extremities of the child’s body, notably in his hair. As a result, the hair stands on end, because the excess negative charge on each strand repels the excess positive charge on neighboring strands. Polarization can be used to charge objects. Consider the two metallic spheres shown in Figure 18.11. The spheres are electrically neutral, so they carry the same amounts of positive and negative charge. In the top picture (Figure 18.11(a)), the two spheres are touching, and the positive and negative charge is evenly distributed over the two spheres. We then approach a glass rod that carries an excess positive charge, which can be done by rubbing the glass rod with silk, as shown in Figure 18.11(b). Because opposite charges attract each other, the negative charge is attracted to the glass rod, leaving an excess positive charge on the opposite side of the right sphere. This is an example of charging by
induction, whereby a charge is created by approaching a charged object with a second object to create an unbalanced charge in the second object. If we then separate the two spheres, as shown in Figure 18.11(c), the excess charge is stuck on each sphere. The left sphere now has an excess negative charge, and the right sphere has an excess positive charge. Finally, in the bottom picture, the rod is removed, and the opposite charges attract each other, so they move as close together as they can get. 558 Chapter 18 • Static Electricity Figure 18.11 (a) Two neutral conducting spheres are touching each other, so the charge is evenly spread over both spheres. (b) A positively charged rod approaches, which attracts negative charges, leaving excess positive charge on the right sphere. (c) The spheres are separated. Each sphere now carries an equal magnitude of excess charge. (d) When the positively charged rod is removed, the excess negative charge on the left sphere is attracted to the excess positive charge on the right sphere. FUN IN PHYSICS Create a Spark in a Science Fair Van de Graaff generators are devices that are used not only for serious physics research but also for demonstrating the physics of static electricity at science fairs and in classrooms. Because they deliver relatively little electric current, they can be made safe for use in such environments. The first such generator was built by Robert Van de Graaff in 1931 for use in nuclear physics research. Figure 18.12 shows a simplified sketch of a Van de Graaff generator. Van de Graaff generators use smooth and pointed surfaces and conductors and insulators to generate large static charges. In the version shown in Figure 18.12, electrons are “sprayed” from the tips of the lower comb onto a moving belt, which is made of an insulating material like, such as rubber. This technique of charging the belt is akin to charging your shoes with electrons by walking across a carpet. The belt raises the charges up to the upper comb, where they transfer again, akin to your touching the doorknob and transferring your charge to it. Because like charges repel, the excess electrons all rush to the outer surface of the globe, which is made of metal (a conductor). Thus, the comb itself never accumulates too much charge, because any charge it gains is quickly depleted by the charge moving to the outer surface of the globe. Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and
Transfer of Charge 559 Figure 18.12 Van de Graaff generators transfer electrons onto a metallic sphere, where the electrons distribute themselves uniformly over the outer surface. Van de Graaff generators are used to demonstrate many interesting effects caused by static electricity. By touching the globe, a person gains excess charge, so his or her hair stands on end, as shown in Figure 18.13. You can also create mini lightning bolts by moving a neutral conductor toward the globe. Another favorite is to pile up aluminum muffin tins on top of the uncharged globe, then turn on the generator. Being made of conducting material, the tins accumulate excess charge. They then repel each other and fly off the globe one by one. A quick Internet search will show many examples of what you can do with a Van de Graaff generator. Figure 18.13 The man touching the Van de Graaff generator has excess charge, which spreads over his hair and repels hair strands from his neighbors. (credit: Jon “ShakataGaNai” Davis) GRASP CHECK Why don’t the electrons stay on the rubber belt when they reach the upper comb? a. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. b. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. c. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. d. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. Virtual Physics Balloons and Static Electricity Click to view content (http://www.openstax.org/l/28balloons) 560 Chapter 18 • Static Electricity This simulation allows you to observe negative charge accumulating on a balloon as you rub it against a sweater. You can then observe how two charged balloons interact and how they cause polarization in a wall. GRASP CHECK Click the reset button, and start with two balloons. Charge a first balloon by rubbing it on the sweater, and then move it toward the second balloon. Why does the second balloon not move? a. The second balloon has an equal number of positive and negative charges. b. The second balloon has more positive charges than negative charges. c. The second balloon has more negative charges than positive charges. d. The second balloon is positively charged and has polarization.
Snap Lab Polarizing Tap Water This lab will demonstrate how water molecules can easily be polarized. • Plastic object of small dimensions, such as comb or plastic stirrer • Source of tap water Instructions Procedure 1. Thoroughly rub the plastic object with a dry cloth. 2. Open the faucet just enough to let a smooth filament of water run from the tap. 3. Move an edge of the charged plastic object toward the filament of running water. What do you observe? What happens when the plastic object touches the water filament? Can you explain your observations? GRASP CHECK Why does the water curve around the charged object? a. The charged object induces uniform positive charge on the water molecules. b. The charged object induces uniform negative charge on the water molecules. c. The charged object attracts the polarized water molecules and ions that are dissolved in the water. d. The charged object depolarizes the water molecules and the ions dissolved in the water. WORKED EXAMPLE Charging Ink Droplets Electrically neutral ink droplets in an ink-jet printer pass through an electron beam created by an electron gun, as shown in Figure 18.14. Some electrons are captured by the ink droplet, so that it becomes charged. After passing through the electron. How many electrons are captured by the ink droplet? beam, the net charge of the ink droplet is Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 561 Figure 18.14 Electrons from an electron guncharge a passing ink droplet. STRATEGY A single electron carries a charge of of a single electron will give the number of electrons captured by the ink droplet.. Dividing the net charge of the ink droplet by the charge Solution The number nof electrons captured by the ink droplet are 18.4 Discussion This is almost a billion electrons! It seems like a lot, but it is quite small compared to the number of atoms in an ink droplet, which number about Thus, each extra electron is shared between about atoms. Practice Problems 3. How many protons are needed to make 1 nC of charge? 1 nC = 10−9 C 1.6 × 10−28 a. 1.6 × 10−10 b. 3 × 109 c. d. 6 × 109 4. In a physics lab, you charge up three metal spheres, two with and one with. When you bring all three spheres together
so that they all touch one another, what is the total charge on the three spheres? a. b. c. d. Check Your Understanding 5. How many types of electric charge exist? a. one type two types b. three types c. four types d. 6. Which are the two main electrical classifications of materials based on how easily charges can move through them? 562 Chapter 18 • Static Electricity a. b. c. d. conductor and insulator semiconductor and insulator conductor and superconductor conductor and semiconductor 7. True or false—A polarized material must have a nonzero net electric charge. a. b. true false 8. Describe the force between two positive point charges that interact. a. The force is attractive and acts along the line joining the two point charges. b. The force is attractive and acts tangential to the line joining the two point charges. c. The force is repulsive and acts along the line joining the two point charges. d. The force is repulsive and acts tangential to the line joining the two point charges. 9. How does a conductor differ from an insulator? a. Electric charges move easily in an insulator but not in a conducting material. b. Electric charges move easily in a conductor but not in an insulator. c. A conductor has a large number of electrons. d. More charges are in an insulator than in a conductor. 10. True or false—Charging an object by polarization requires touching it with an object carrying excess charge. a. b. true false 18.2 Coulomb's law Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Coulomb’s law verbally and mathematically • Solve problems involving Coulomb’s law Section Key Terms Coulomb’s law inverse-square law More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist Charles-Augustin de Coulomb mathematically described the force between charged objects. Doing so required careful measurements of forces between charged spheres, for which he built an ingenious device called a torsion balance. This device, shown in Figure 18.15, contains an insulating rod that is hanging by a thread inside a glass-walled enclosure. At one end of the rod is the metallic sphere A. When no charge is on this sphere, it touches sphere B. Coulomb would touch the spheres with a third
metallic ball (shown at the bottom of the diagram) that was charged. An unknown amount of charge would distribute evenly between spheres A and B, which would then repel each other, because like charges repel. This force would cause sphere A to rotate away from sphere B, thus twisting the wire until the torsion in the wire balanced the electrical force. Coulomb then turned the knob at the top, which allowed him to rotate the thread, thus bringing sphere A closer to sphere B. He found that bringing sphere A twice as close to sphere B required increasing the torsion by a factor of four. Bringing the sphere three times closer required a ninefold increase in the torsion. From this type of measurement, he deduced that the electrical force between the spheres was inversely proportional to the distance squared between the spheres. In other words, where ris the distance between the spheres. An electrical charge distributes itself equally between two conducting spheres of the same size. Knowing this allowed Coulomb to divide an unknown charge in half. Repeating this process would produce a sphere with one quarter of the initial charge, and 18.5 Access for free at openstax.org. so on. Using this technique, he measured the force between spheres A and B when they were charged with different amounts of charge. These measurements led him to deduce that the force was proportional to the charge on each sphere, or 18.6 where is the charge on sphere A, and is the charge on sphere B. 18.2 • Coulomb's law 563 Figure 18.15 A drawing of Coulomb’s torsion balance, which he used to measure the electrical force between charged spheres. (credit: Charles-Augustin de Coulomb) Combining these two proportionalities, he proposed the following expression to describe the force between the charged spheres. This equation is known as Coulomb’s law, and it describes the electrostatic force between charged objects. The constant of proportionality kis called Coulomb’s constant. In SI units, the constant khas the value 18.7 The direction of the force is along the line joining the centers of the two objects. If the two charges are of opposite signs, Coulomb’s law gives a negative result. This means that the force between the particles is attractive. If the two charges have the same signs, Coulomb’s law gives a positive result. This means that the force between the particles is repulsive. For example, if
and both negative charge and This is shown in Figure 18.16(b). is a is a positive charge (or vice versa), then the charges are different, so the force between them is attractive. are negative or if both are positive, the force between them is repulsive. This is shown in Figure 18.16(a). If Figure 18.16 The magnitude of the electrostatic force Fbetween point charges q1 and q2 separated by a distance ris given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force (F1,2) on q1 is equal in magnitude and opposite in direction to the force (F2,1) it exerts on q2. (a) Like charges. (b) Unlike charges. Note that Coulomb’s law applies only to charged objects that are not moving with respect to each other. The law says that the force is proportional to the amount of charge on each object and inversely proportional to the square of the distance between the objects. If we double the charge the force between them decreasesby a factor of spherical objects or to objects that are much smaller than the distance between the objects (in which case, the objects can be approximated as spheres)., for instance, then the force is doubled. If we double the distance between the objects, then. Although Coulomb’s law is true in general, it is easiest to apply to 564 Chapter 18 • Static Electricity Coulomb’s law is an example of an inverse-square law, which means the force depends on the square of the denominator. Another inverse-square law is Newton’s law of universal gravitation, which is they differ in two important respects: (i) The gravitational constant Gis much, much smaller than k ( differences explain why gravity is so much weaker than the electrostatic force and why gravity is only attractive, whereas the electrostatic force can be attractive or repulsive. ); and (ii) only one type of mass exists, whereas two types of electric charge exist. These two. Although these laws are similar, Finally, note that Coulomb measured the distance between the spheres from the centers of each sphere. He did not explain this assumption in his original papers, but it turns out to be valid. From outsidea uniform spherical distribution of charge, it can be treated as if all the charge were located at the center of the sphere. WATCH
PHYSICS Electrostatics (part 1): Introduction to charge and Coulomb's law This video explains the basics of Coulomb’s law. Note that the lecturer uses dfor the distance between the center of the particles instead of r. Click to view content (https://www.openstax.org/l/28coulomb) GRASP CHECK True or false—If one particle carries a positive charge and another carries a negative charge, then the force between them is attractive. true a. false b. Snap Lab Hovering plastic In this lab, you will use electrostatics to hover a thin piece of plastic in the air. • Balloon • Light plastic bag (e.g., produce bag from grocery store) Instructions Inflate the balloon. Procedure 1. Cut the plastic bag to make a plastic loop about 2 inches wide. 2. 3. Charge the balloon by rubbing it on your clothes. 4. Charge the plastic loop by placing it on a nonmetallic surface and rubbing it with a cloth. 5. Hold the balloon in one hand, and in the other hand hold the plastic loop above the balloon. If the loop clings too much to your hand, recruit a friend to hold the strip above the balloon with both hands. Now let go of the plastic loop, and maneuver the balloon under the plastic loop to keep it hovering in the air above the balloon. GRASP CHECK How does the balloon keep the plastic loop hovering? a. The balloon and the loop are both negatively charged. This will help the balloon keep the plastic loop hovering. b. The balloon is charged, while the plastic loop is neutral.This will help the balloon keep the plastic loop hovering. c. The balloon and the loop are both positively charged. This will help the balloon keep the plastic loop hovering. d. The balloon is positively charged, while the plastic loop is negatively charged. This will help the balloon keep the plastic loop hovering. Access for free at openstax.org. WORKED EXAMPLE 18.2 • Coulomb's law 565 Using Coulomb’s law to find the force between charged objects between the two charged spheres when they are separated by 5.0 cm. By Suppose Coulomb measures a force of turning the dial at the top of the torsion balance, he approaches the spheres so that they are separated by 3.0 cm. Which force does he measure now? STRATEGY Apply Coulomb’s law to the situation before and
after the spheres are brought closer together. Although we do not know the charges on the spheres, we do know that they remain the same. We call these unknown but constant charges Because these charges appear as a product in Coulomb’s law, they form a single unknown. We thus have two equations and two unknowns, which we can solve. The first unknown is the force (which we call second is ) when the spheres are 3.0 cm apart, and the and.. Use the following notation: When the charges are 5.0 cm apart, the force is where the subscript imeans initial. Once the charges are brought closer together, we know the subscript fmeans final. and,, where Solution Coulomb’s law applied to the spheres in their initial positions gives Coulomb’s law applied to the spheres in their final positions gives Dividing the second equation by the first and solving for the final force leads to Inserting the known quantities yields 18.8 18.9 18.10 18.11 The force acts along the line joining the centers of the spheres. Because the same type of charge is on each sphere, the force is repulsive. Discussion As expected, the force between the charges is greater when they are 3.0 cm apart than when they are 5.0 cm apart. Note that although it is a good habit to convert cm to m (because the constant kis in SI units), it is not necessary in this problem, because the distances cancel out. We can also solve for the second unknown. By using the first equation, we find 566 Chapter 18 • Static Electricity 18.12 Note how the units cancel in the second-to-last line. Had we not converted cm to m, this would not occur, and the result would be incorrect. Finally, because the charge on each sphere is the same, we can further deduce that 18.13 WORKED EXAMPLE Using Coulomb’s law to find the distance between charged objects An engineer measures the force between two ink drops by measuring their acceleration and their diameter. She finds that each member of a pair of ink drops exerts a repulsive force of on its partner. If each ink drop carries a charge, how far apart are the ink drops? STRATEGY We know the force and the charge on each ink drop, so we can solve Coulomb’s law for the distance rbetween the ink drops. Do not forget to convert the force into SI units: Solution
The charges in Coulomb’s law are so the numerator in Coulomb’s law takes the form. Inserting this into Coulomb’s law and solving for the distance rgives 18.14 or 130 microns (about one-tenth of a millimeter). Discussion The plus-minus sign means that we do not know which ink drop is to the right and which is to the left, but that is not important, because both ink drops are the same. Practice Problems 11. A charge of −4 × 10−9 C is a distance of 3 cm from a charge of 3 × 10−9 C. What is the magnitude and direction of the force between them? 1.2 × 10−4 N, and the force is attractive a. 1.2 × 1014 N, and the force is attractive b. c. 6.74 × 1023 N, and the force is attractive d. −ŷ, and the force is attractive 12. Two charges are repelled by a force of 2.0 N. If the distance between them triples, what is the force between the charges? a. 0.22 N b. 0.67 N c. 2.0 N 18.0 N d. Access for free at openstax.org. 18.3 • Electric Field 567 Check Your Understanding 13. How are electrostatic force and charge related? a. The force is proportional to the product of two charges. b. The force is inversely proportional to the product of two charges. c. The force is proportional to any one of the charges between which the force is acting. d. The force is inversely proportional to any one of the charges between which the force is acting. 14. Why is Coulomb’s law called an inverse-square law? a. because the force is proportional to the inverse of the distance squared between charges b. because the force is proportional to the product of two charges c. because the force is proportional to the inverse of the product of two charges d. because the force is proportional to the distance squared between charges 18.3 Electric Field Section Learning Objectives By the end of this section, you will be able to do the following: • Calculate the strength of an electric field • Create and interpret drawings of electric fields Section Key Terms electric field test charge You may have heard of a force fieldin science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect
a spaceship from enemy fire. The concept of a fieldis very useful in physics, although it differs somewhat from what you see in movies. A fieldis a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field. Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field. If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field. An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution. The charge distribution could be a single point charge; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field. A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field. The electric field exerts a force on the test charge in a given direction. The force exerted is proportional to the charge of the test charge. For example, if we double the charge of the test charge, the force exerted on it doubles. Mathematically, saying that electric field is the force per unit charge is written as 18.15 where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge. The units of electric field are N/C. If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Qand the test charge is given by Coulomb’s law where the absolute value is used, because we only consider the magnitude of the force. The magnitude of the electric field is then 568 Chapter 18 • Static Electricity 18.16 This equation gives the magnitude of the electric field created by a point charge Q. The distance rin the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest. If the test charge is removed from the electric field, the electric field still exists. To create a three-dimensional map of the electric field, imagine
placing the test charge in various locations in the field. At each location, measure the force on the charge, and use the vector equation to calculate the electric field. Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure 18.17 shows an image of the threedimensional electric field created by a positive charge. Figure 18.17 Three-dimensional representation of the electric field generated by a positive charge. Just drawing the electric field lines in a plane that slices through the charge gives the two-dimensional electric-field maps shown in Figure 18.18. On the left is the electric field created by a positive charge, and on the right is the electric field created by a negative charge. Notice that the electric field lines point away from the positive charge and toward the negative charge. Thus, a positive test charge placed in the electric field of the positive charge will be repelled. This is consistent with Coulomb’s law, which says that like charges repel each other. If we place the positive charge in the electric field of the negative charge, the positive charge is attracted to the negative charge. The opposite is true for negative test charges. Thus, the direction of the electric field lines is consistent with what we find by using Coulomb’s law. says that the electric field gets stronger as we approach the charge that generates it. For example, at The equation 2 cm from the charge Q(r= 2 cm), the electric field is four times stronger than at 4 cm from the charge (r= 4 cm). Looking at Figure 18.17 and Figure 18.18 again, we see that the electric field lines become denser as we approach the charge that generates it. In fact, the density of the electric field lines is proportional to the strength of the electric field! Figure 18.18 Electric field lines from two point charges. The red point on the left carries a charge of +1 nC, and the blue point on the right carries a charge of –1 nC. The arrows point in the direction that a positive test charge would move. The field lines are denser as you approach the point charge. Electric-field maps can be made for several charges or for more complicated charge distributions. The electric field due to multiple charges may be found by adding together the electric field from each individual charge. Because
this sum can only be a single number, we know that only a single electric-field line can go through any given point. In other words, electric-field lines cannotcross each other. Figure 18.19(a) shows a two-dimensional map of the electric field generated by a charge of +qand a nearby charge of −q. The three-dimensional version of this map is obtained by rotating this map about the axis that goes through both charges. A positive Access for free at openstax.org. test charge placed in this field would experience a force in the direction of the field lines at its location. It would thus be repelled from the positive charge and attracted to the negative charge. Figure 18.19(b) shows the electric field generated by two charges of −q. Note how the field lines tend to repel each other and do not overlap. A positive test charge placed in this field would be attracted to both charges. If you are far from these two charges, where far means much farther than the distance between the charges, the electric field looks like the electric field from a single charge of −2q. 18.3 • Electric Field 569 Figure 18.19 (a) The electric field generated by a positive point charge (left) and a negative point charge of the same magnitude (right). (b) The electric field generated by two equal negative charges. Virtual Physics Probing an Electric Field Click to view content (http://www.openstax.org/l/28charge-field) This simulation shows you the electric field due to charges that you place on the screen. Start by clicking the top checkbox in the options panel on the right-hand side to show the electric field. Drag charges from the buckets onto the screen, move them around, and observe the electric field that they form. To see more precisely the magnitude and direction of the electric field, drag an electric-field sensor, or E-fieldsensor from the bottom bucket, and move it around the screen. GRASP CHECK Two positive charges are placed on a screen. Which statement describes the electric field produced by the charges? a. b. c. d. It is constant everywhere. It is zero near each charge. It is zero halfway between the charges. It is strongest halfway between the charges. WATCH PHYSICS Electrostatics (part 2): Interpreting electric field This video explains how to calculate the electric field of a point charge and how to interpret electric-field maps
in general. Note 570 Chapter 18 • Static Electricity that the lecturer uses dfor the distance between particles instead of r. Note that the point charges are infinitesimally small, so all their charges are focused at a point. When larger charged objects are considered, the distance between the objects must be measured between the center of the objects. Click to view content (https://www.youtube.com/embed/0YOGrTNgGhE) GRASP CHECK True or false—If a point charge has electric field lines that point into it, the charge must be ositive. a. b. true false WORKED EXAMPLE What is the charge? Look at the drawing of the electric field in Figure 18.20. What is the relative strength and sign of the three charges? Figure 18.20 Map of electric field due to three charged particles. STRATEGY We know the electric field extends out from positive charge and terminates on negative charge. We also know that the number of electric field lines that touch a charge is proportional to the charge. Charge 1 has 12 fields coming out of it. Charge 2 has six field lines going into it. Charge 3 has 12 field lines going into it. Solution The electric-field lines come out of charge 1, so it is a positive charge. The electric-field lines go into charges 2 and 3, so they are negative charges. The ratio of the charges is of charge 2.. Thus, magnitude of charges 1 and 3 is twice that Discussion Although we cannot determine the precise charge on each particle, we can get a lot of information from the electric field regarding the magnitude and sign of the charges and where the force on a test charge would be greatest (or least). WORKED EXAMPLE Electric field from doorknob A doorknob, which can be taken to be a spherical metal conductor, acquires a static electricity charge of the electric field 1.0 cm in front of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a to indicate the outward direction later physics course. Now sketch the
doorknob, and define your coordinate system. Use What is Access for free at openstax.org. perpendicular to the door, with at the center of the doorknob (as shown in the figure below). 18.3 • Electric Field 571 If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance from the center of the doorknob. We can use the equation to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case. Solution Inserting the charge gives and the distance into the equation 18.17 Because the charge is negative, the electric-field lines point toward the center of the doorknob. Thus, the electric field at is. Discussion This seems like an enormous electric field. Luckily, it takes an electric field roughly 100 times stronger ( air to break down and conduct electricity. Also, the weight of an adult is about feel a force on the protons in your hand as you reach for the doorknob? The reason is that your hand contains an equal amount of negative charge, which repels the negative charge in the doorknob. A very small force might develop from polarization in your hand, but you would never notice it. ) to cause, so why don’t you Practice Problems 15. What is the magnitude of the electric field from 20 cm from a point charge of q= 33 nC? 7.4 × 103 N/C 1.48 × 103 N / C 7.4 × 1012 N / C a. b. c. d. 0 16. A −10 nC charge is at the origin. In which direction does the electric field from the charge point at x+ 10 cm? a. The electric field points away from negative charges. b. The electric field points toward negative charges. c. The electric field points toward positive charges. d. The electric field points away from positive charges. Check Your Understanding 17. When electric field lines get closer together, what does that tell you about the electric field? 572 Chapter 18 • Static Electricity a. The electric field is inversely proportional to the density of electric field lines. b. The electric field is directly proportional to the density of electric field lines. c. The electric field is not related to the density of electric field lines
. d. The electric field is inversely proportional to the square root of density of electric field lines. 18. If five electric-field lines come out of a +5 nC charge, how many electric-field lines should come out of a +20 nC charge? a. five field lines 10 field lines b. c. 15 field lines d. 20 field lines 18.4 Electric Potential Section Learning Objectives By the end of this section, you will be able to do the following: • Explain the similarities and differences between electric potential energy and gravitational potential energy • Calculate the electric potential difference between two point charges and in a uniform electric field Section Key Terms electric potential electric potential energy As you learned in studying gravity, a mass in a gravitational field has potential energy, which means it has the potential to accelerate and thereby increase its kinetic energy. This kinetic energy can be used to do work. For example, imagine you want to use a stone to pound a nail into a piece of wood. You first lift the stone high above the nail, which increases the potential energy of the stone-Earth system—because Earth is so large, it does not move, so we usually shorten this by saying simply that the potential energy of the stone increases. When you drop the stone, gravity converts the potential energy into kinetic energy. When the stone hits the nail, it does work by pounding the nail into the wood. The gravitational potential energy is the work that a mass can potentially do by virtue of its position in a gravitational field. Potential energy is a very useful concept, because it can be used with conservation of energy to calculate the motion of masses in a gravitational field. Electric potential energy works much the same way, but it is based on the electric field instead of the gravitational field. By virtue of its position in an electric field, a charge has an electric potential energy. If the charge is free to move, the force due to the electric field causes it to accelerate, so its potential energy is converted to kinetic energy, just like a mass that falls in a gravitational field. This kinetic energy can be used to do work. The electric potential energy is the work that a charge can do by virtue of its position in an electric field. The analogy between gravitational potential energy and electric potential energy is depicted in Figure 18.21. On the left, the ballEarth system gains gravitational potential energy when the ball is higher in Earth's gravitational field. On the right, the twocharge system gains electric potential energy when the positive charge is farther from the
negative charge. Access for free at openstax.org. 18.4 • Electric Potential 573 Figure 18.21 On the left, the gravitational field points toward Earth. The higher the ball is in the gravitational field, the higher the potential energy is of the Earth-ball system. On the right, the electric field points toward a negative charge. The farther the positive charge is from the negative charge, the higher the potential energy is of the two-charge system. Let’s use the symbol potential energy decreases. Conservation of energy tells us that the work done by the gravitational field to make the mass accelerate must equal the loss of potential energy of the mass. If we use the symbol to denote gravitational potential energy. When a mass falls in a gravitational field, its gravitational to denote this work, then where the minus sign reflects the fact that the potential energy of the ball decreases. The work done by gravity on the mass is 18.18 18.19 where Fis the force due to gravity, and are the initial and final positions of the ball, respectively. The negative sign is because gravity points down, which we consider to be the negative direction. For the constantgravitational field near Earth’s surface,. The change in gravitational potential energy of the mass is and 18.20. in electric Note that is just the negative of the height hfrom which the mass falls, so we usually just write We now apply the same reasoning to a charge in an electric field to find the electric potential energy. The change potential energy is the work done by the electric field to move a charge qfrom an initial position ( ). The definition of work does not change, except that now the work is done by the electric field: to a final position electric field is. The change in electric potential energy of the charge is thus. For a charge that falls through a constantelectric field E, the force applied to the charge by the or 18.21 18.22 This equation gives the change in electric potential energy of a charge qwhen it moves from position constantelectric field E. to position in a Figure 18.22 shows how this analogy would work if we were close to Earth’s surface, where gravity is constant. The top image shows a charge accelerating due to a constant electric field. Likewise, the round mass in the bottom image accelerates due to a constant gravitation field. In both cases, the potential energy of the particle decreases, and its kinetic energy increases. 574 Chapter 18 • Static Electricity Figure
18.22 In the top picture, a mass accelerates due to a constant electric field. In the bottom picture, the mass accelerates due to a constant gravitational field. WATCH PHYSICS Analogy between Gravity and Electricity This video discusses the analogy between gravitational potential energy and electric potential energy. It reviews the concepts of work and potential energy and shows the connection between a mass in a uniform gravitation field, such as on Earth’s surface, and an electric charge in a uniform electric field. Click to view content (https://www.openstax.org/l/28grav-elec) If the electric field is not constant, then the equation energy becomes more involved. For example, consider the electric potential energy of an assembly of two point charges is not valid, and deriving the electric potential and of the same sign that are initially very far apart. We start by placing charge at the origin of our coordinate system. This in from very far away to a distance rfrom the center of charge takes no electrical energy, because there is no electric field at the origin (because charge charge of charge. The energy it takes to assemble these two charges can be recuperated if we let them fly apart again. Thus, the charges have potential energy when they are a distance rapart. It turns out that the electric a distance rapart is potential energy of a pair of point charges. This requires some effort, because the electric field applies a repulsive force on charge is very far away). We then bring and To recap, if charges are free to move, they can accumulate kinetic energy by flying apart, and this kinetic energy can be used to do work. The maximum amount of work the two charges can do (if they fly infinitely far from each other) is given by the equation above. and Notice that if the two charges have opposite signs, then the potential energy is negative. This means that the charges have more potential to do work when they are farapart than when they are at a distance rapart. This makes sense: Opposite charges attract, so the charges can gain more kinetic energy if they attract each other from far away than if they start at only a short distance apart. Thus, they have more potential to do work when they are far apart. Figure 18.23 summarizes how the electric potential energy depends on charge and separation. 18.23 Access for free at openstax.org. 18.4 • Electric Potential 575 Figure 18.23 The potential energy depends on the sign of
the charges and their separation. The arrows on the charges indicate the direction in which the charges would move if released. When charges with the same sign are far apart, their potential energy is low, as shown in the top panel for two positive charges. The situation is the reverse for charges of opposite signs, as shown in the bottom panel. Electric Potential Recall that to find the force applied by a fixed charge Qon any arbitrary test charge q, it was convenient to define the electric field, which is the force per unit charge applied by Qon any test charge that we place in its electric field. The same strategy is used here with electric potential energy: We now define the electric potential V, which is the electric potential energy per unit charge. 18.24 Normally, the electric potential is simply called the potentialor voltage. The units for the potential are J/C, which are given the name volt(V) after the Italian physicist Alessandro Volta (1745–1827). From the equation distance rfrom a point charge, the electric potential a is 18.25 This equation gives the energy required per unit charge to bring a charge Mathematically, this is written as from infinity to a distance rfrom a point charge ∞ 18.26 Note that this equation actually represents a differencein electric potential. However, because the second term is zero, it is normally not written, and we speak of the electric potential instead of the electric potential difference, or we just say the potential difference, or voltage). Below, when we consider the electric potential energy per unit charge between two points not infinitely far apart, we speak of electric potential differenceexplicitly. Just remember that electric potential and electric potential difference are really the same thing; the former is used just when the electric potential energy is zero in either the initial or final charge configuration. Coming back now to the electric potential a distance rfrom a point charge we can drop the subscripts and simply write, note that can be any arbitrary point charge, so Now consider the electric potential near a group of charges q1, q2, and q3, as drawn in Figure 18.24. The electric potential is 18.27 576 Chapter 18 • Static Electricity derived by considering the electric field. Electric fields follow the principle of superposition and can be simply added together, so the electric potential from different charges also add together. Thus, the electric potential of a point near a group of charges is where 18.24. are the distances from the center of charges to the point of interest
, as shown in Figure 18.28 Figure 18.24 The potential at the red point is simply the sum of the potentials due to each individual charge. Now let’s consider the electric potential in a uniform electric field. From the equation potential difference in going from to in a uniform electric field Eis, we see that the TIPS FOR SUCCESS Notice from the equation that the electric field can be written as 18.29 18.30 which means that the electric field has units of V/m. Thus, if you know the potential difference between two points, calculating the electric field is very simple—you simply divide the potential difference by the distance! Notice that a positive charge in a region with high potential will experience a force pushing it toward regions of lower potential. In this sense, potential is like pressure for fluids. Imagine a pipe containing fluid, with the fluid at one end of the pipe under high pressure and the fluid at the other end of the pipe under low pressure. If nothing prevents the fluid from flowing, it will flow from the high-pressure end to the low-pressure end. Likewise, a positive charge that is free to move will move from a region with high potential to a region with lower potential. WATCH PHYSICS Voltage This video starts from electric potential energy and explains how this is related to electric potential (or voltage). The lecturer calculates the electric potential created by a uniform electric field. Click to view content (https://www.openstax.org/l/28voltage) GRASP CHECK What is the voltage difference between the positions a. and in an electric field of? Access for free at openstax.org. b. c. d. 18.4 • Electric Potential 577 LINKS TO PHYSICS Electric Animals Many animals generate and/or detect electric fields. This is useful for activities such as hunting, defense, navigation, communication, and mating. Because salt water is a relatively good conductor, electric fish have evolved in all the world’s oceans. These fish have intrigued humans since the earliest times. In the nineteenth century, parties were even organized where the main attraction was getting a jolt from an electric fish! Scientists also studied electric fish to learn about electricity. Alessandro Volta based his research that led to batteries in 1799 on electric fish. He even referred to batteries as artificial electric organs, because he saw them as imitations of the electric organs of electric fish. Animals that generate electricity are called electrogenicand those that
detect electric fields are called electroreceptive. Most fish that are electrogenic are also electroreceptive. One of the most well-known electric fish is the electric eel (see Figure 18.25), which is both electrogenic and electroreceptive. These fish have three pairs of organs that produce the electric charge: the main organ, Hunter’s organ, and Sach’s organ. Together, these organs account for more than 80percent of the fish’s body. Electric eels can produce electric discharges of much greater voltage than what you would get from a standard wall socket. These discharges can stun or even kill their prey. They also use low-intensity discharges to navigate. The electric fields they generate reflect off nearby obstacles or animals and are then detected by electroreceptors in the eel’s skin. The three organs that produce electricity contain electrolytes, which are substances that ionize when dissolved in water (or other liquids). An ionized atom or molecule is one that has lost or gained at least one electron, so it carries a net charge. Thus, a liquid solution containing an electrolyte conducts electricity, because the ions in the solution can move if an electric field is applied. To produce large discharges, the main organ is used. It contains approximately 6,000 rows of electroplaques connected in a long chain. Connected this way, the voltage between electroplaques adds up, creating a large final voltage. Each electroplaque consists of a column of cells controlled by an excitor nerve. When triggered by the excitor nerve, the electroplaques allow ionized sodium to flow through them, creating a potential difference between electroplaques. These potentials add up, and a large current can flow through the electrolyte. This geometry is reflected in batteries, which also use stacks of plates to produce larger potential differences. Figure 18.25 An electric eel in its natural environment. (credit: Steven G. Johnson) GRASP CHECK If an electric eel produces 1,000 V, which voltage is produced by each electroplaque in the main organ? a. 0.17 mV b. c. d. 1.7 mV 17 mV 170 mV 578 Chapter 18 • Static Electricity WORKED EXAMPLE X-ray Tube Dentists use X-rays to image their patients’ teeth and bones. The X-ray tubes that generate X-rays contain an electron source separated by about 10
cm from a metallic target. The electrons are accelerated from the source to the target by a uniform electric field with a magnitude of about 100 kN/C, as drawn in Figure 18.26. When the electrons hit the target, X-rays are produced. (a) What is the potential difference between the electron source and the metallic target? (b) What is the kinetic energy of the electrons when they reach the target, assuming that the electrons start at rest? Figure 18.26 In an X-ray tube, a large current flows through the electron source, causing electrons to be ejected from the electron source. The ejected electrons are accelerated toward the target by the electric field. When they strike the target, X-rays are produced. STRATEGY FOR (A) Use the equation as point in the negative xdirection. This way, the force qand Eare negative. Thus, and the target position as. to find the potential difference given a constant electric field. Define the source position. To accelerate the electrons in the positive xdirection, the electric field must on the electrons will point in the positive xdirection, because both Solution for (a) and Using electron source and the target is, the equation tells us that the potential difference between the Discussion for (a) The potential difference is positive, so the energy per unit positive charge is higher at the target than at the source. This means that free positive charges would fall from the target to the source. However, electrons are negative charges, so they accelerate from the source toward the target, gaining kinetic energy as they go. STRATEGY FOR (B) Apply conservation of energy to find the final kinetic energy of the electrons. In going from the source to the target, the change The change in in electric potential energy plus the change in kinetic energy of the electrons must be zero, so electric potential energy for moving through a constant electric field is given by the equation 18.31 where the electric field is the change in kinetic energy is simply their final kinetic energy, so.. Because the electrons start at rest, their initial kinetic energy is zero. Thus, Solution for (b) Again gives and. The charge of an electron is. Conservation of energy 18.32 Access for free at openstax.org. Inserting the known values into the right-hand side of this equation gives 18.4 • Electric Potential 579 18.33 Discussion for (b) This is a very small energy. However, electrons are very small, so they are easy to
accelerate, and this energy is enough to make. The result is that an electron go extremely fast. You can find their speed by using the definition of kinetic energy, the electrons are moving at more than 100 million miles per hour! WORKED EXAMPLE Electric Potential Energy of Doorknob and Dust Speck Consider again the doorknob from the example in the previous section. The doorknob is treated as a spherical conductor with a on its surface. What is the electric potential energy between the doorknob and a speck of uniform static charge dust carrying a charge at 1.0 cm from the front surface of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY As we did in the previous section, we treat the charge as if it were concentrated at the center of the doorknob. Again, as you will be able to validate in later physics classes, we can make this simplification, because the charge is uniformly distributed over the surface of the spherical object. Make a sketch of the situation and define a coordinate system, as shown in the image below. We at the center of the doorknob. If the diameter of use the doorknob is 5.0 cm, its radius is 2.5 cm. Thus, the speck of dust 1.0 cm from the surface of the doorknob is a distance to indicate the outward direction perpendicular to the door, with from the center of the doorknob. To solve this problem, use the equation. Solution The charge on the doorknob is gives. The distance. Inserting these values into the equation and the charge on the speck of dust is Discussion The energy is negative, which means that the energy will decrease that is, get even morenegative as the speck of dust approaches the doorknob. This helps explain why dust accumulates on objects that carry a static charge. However, note that insulators normally collect more static charge than conductors, because any charge that accumulates on insulators cannot move about on the insulator to find a way to escape. They must simply wait to be removed by some passing moist speck of dust or other host. 18.34 580 Chapter 18 • Static Electricity Practice Problems 19. What is the electric potential 10 cm from a −10 nC charge? a. 9.0 × 102 V b. 9.0 × 103 V c. 9.0 × 104 V d. 9.0 × 105 V 20. An electron accelerates
from 0 to 10 × 104 m/s in an electric field. Through what potential difference did the electron travel? The mass of an electron is 9.11 × 10–31 kg, and its charge is −1.60 × 10–19 C. a. 29 mV b. 290 mV c. 2,900 mV d. 29 V Check Your Understanding 21. Gravitational potential energy is the potential for two masses to do work by virtue of their positions with respect to each other. What is the analogous definition of electric potential energy? a. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to the origin point. b. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to infinity. c. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to each other. d. Electric potential energy is the potential for single charges to do work by virtue of their positions with respect to their final positions. 22. A negative charge is 10 m from a positive charge. Where would you have to move the negative charge to increase the potential energy of the system? a. The negative charge should be moved closer to the positive charge. b. The negative charge should be moved farther away from the positive charge. c. The negative charge should be moved to infinity. d. The negative charge should be placed just next to the positive charge. 18.5 Capacitors and Dielectrics Section Learning Objectives By the end of this section, you will be able to do the following: • Calculate the energy stored in a charged capacitor and the capacitance of a capacitor • Explain the properties of capacitors and dielectrics Section Key Terms capacitor dielectric Capacitors Consider again the X-ray tube discussed in the previous sample problem. How can a uniform electric field be produced? A single positive charge produces an electric field that points away from it, as in. This field is not uniform, because the space between the lines increases as you move away from the charge. However, if we combine a positive and a negative charge, we obtain the electric field shown in (a). Notice that, between the charges, the electric field lines are more equally spaced. What happens if we place, say, five positive charges in a line across from five negative charges, as in Figure 18.27? Now the region between the lines of charge contains a fairly uniform electric field. Access for free
at openstax.org. 18.5 • Capacitors and Dielectrics 581 Figure 18.27 The red dots are positive charges, and the blue dots are negative charges. The electric-field direction is shown by the red arrows. Notice that the electric field between the positive and negative dots is fairly uniform. We can extend this idea even further and into two dimensions by placing two metallic plates face to face and charging one with positive charge and the other with an equal magnitude of negative charge. This can be done by connecting one plate to the positive terminal of a battery and the other plate to the negative terminal, as shown in Figure 18.28. The electric field between these charged plates will be extremely uniform. 582 Chapter 18 • Static Electricity Figure 18.28 Two parallel metal plates are charged with opposite charge, by connecting the plates to the opposite terminals of a battery. The magnitude of the charge on each plate is the same. Let’s think about the work required to charge these plates. Before the plates are connected to the battery, they are neutral—that is, they have zero net charge. Placing the first positive charge on the left plate and the first negative charge on the right plate requires very little work, because the plates are neutral, so no opposing charges are present. Now consider placing a second positive charge on the left plate and a second negative charge on the right plate. Because the first two charges repel the new arrivals, a force must be applied to the two new charges over a distance to put them on the plates. This is the definition of work, which means that, compared with the first pair, more work is required to put the second pair of charges on the plates. To place the third positive and negative charges on the plates requires yet more work, and so on. Where does this work come from? The battery! Its chemical potential energy is converted into the work required to separate the positive and negative charges. Although the battery does work, this work remains within the battery-plate system. Therefore, conservation of energy tells us that, if the potential energy of the battery decreases to separate charges, the energy of another part of the system must increase by the same amount. In fact, the energy from the battery is stored in the electric field between the plates. This idea is analogous to considering that the potential energy of a raised hammer is stored in Earth’s gravitational field. If the gravitational field were to disappear, the hammer would have no potential energy. Likewise, if no
electric field existed between the plates, no energy would be stored between them. If we now disconnect the plates from the battery, they will hold the energy. We could connect the plates to a lightbulb, for example, and the lightbulb would light up until this energy was used up. These plates thus have the capacity to store energy. For this reason, an arrangement such as this is called a capacitor. A capacitor is an arrangement of objects that, by virtue of their geometry, can store energy an electric field. Various real capacitors are shown in Figure 18.29. They are usually made from conducting plates or sheets that are separated by Access for free at openstax.org. an insulating material. They can be flat or rolled up or have other geometries. 18.5 • Capacitors and Dielectrics 583 The capacity of a capacitor is defined by its capacitance C, which is given by Figure 18.29 Some typical capacitors. (credit: Windell Oskay) 18.35 where Qis the magnitudeof the charge on each capacitor plate, and Vis the potential difference in going from the negative plate to the positive plate. This means that both Qand Vare always positive, so the capacitance is always positive. We can see from the equation for capacitance that the units of capacitance are C/V, which are called farads (F) after the nineteenth-century English physicist Michael Faraday. makes sense: A parallel-plate capacitor (like the one shown in Figure 18.28) the size of a football field The equation could hold a lot of charge without requiring too much work per unit charge to push the charge into the capacitor. Thus, Qwould be large, and Vwould be small, so the capacitance Cwould be very large. Squeezing the same charge into a capacitor the size of a fingernail would require much more work, so Vwould be very large, and the capacitance would be much smaller. Although the equation makes it seem that capacitance depends on voltage, in fact it does not. For a given capacitor, the ratio of the charge stored in the capacitor to the voltage difference between the plates of the capacitor always remains the same. Capacitance is determined by the geometry of the capacitor and the materials that it is made from. For a parallel-plate capacitor with nothing between its plates, the capacitance is given by 18.36 where Ais the area of the plates of the capacitor and dis their separation
. We use nothing between its plates (in the next section, we’ll see what happens when this is not the case). The constant zerois called the permittivity of free space, and its value is instead of C, because the capacitor has read epsilon 18.37 Coming back to the energy stored in a capacitor, we can ask exactly how much energy a capacitor stores. If a capacitor is charged by putting a voltage Vacross it for example, by connecting it to a battery with voltage V—the electrical potential energy stored in the capacitor is 18.38 Notice that the form of this equation is similar to that for kinetic energy,. WATCH PHYSICS Where does Capacitance Come From? This video shows how capacitance is defined and why it depends only on the geometric properties of the capacitor, not on voltage or charge stored. In so doing, it provides a good review of the concepts of work and electric potential. Click to view content (https://www.openstax.org/l/28capacitance) 584 Chapter 18 • Static Electricity GRASP CHECK If you increase the distance between the plates of a capacitor, how does the capacitance change? a. Doubling the distance between capacitor plates will reduce the capacitance four fold. b. Doubling the distance between capacitor plates will reduce the capacitance two fold. c. Doubling the distance between capacitor plates will increase the capacitance two times. d. Doubling the distance between capacitor plates will increase the capacitance four times. Virtual Physics Charge your Capacitor Click to view content (http://www.openstax.org/l/28charge-cap) For this simulation, choose the tab labeled Introductionat the top left of the screen. You are presented with a parallel-plate capacitor connected to a variable-voltage battery. The battery is initially at zero volts, so no charge is on the capacitor. Slide the battery slider up and down to change the battery voltage, and observe the charges that accumulate on the plates. Display the capacitance, top-plate charge, and stored energy as you vary the battery voltage. You can also display the electric-field lines in the capacitor. Finally, probe the voltage between different points in this circuit with the help of the voltmeter, and probe the electric field in the capacitor with the help of the electric-field detector. GRASP CHECK True or false— In a capacitor, the stored energy is always positive, regardless of whether the top plate
is charged with negative or positive charge. a. b. false true WORKED EXAMPLE Capacitance and Charge Stored in a Parallel Plate Capacitor (a) What is the capacitance of a parallel-plate capacitor with metal plates, each of area 1.00 m2, separated by 0.0010 m? (b) What charge is stored in this capacitor if a voltage of 3.00 × 103 V is applied to it? STRATEGY FOR (A) Use the equation. Solution for (a) Entering the given values into this equation for the capacitance of a parallel-plate capacitor yields 18.39 Discussion for (a) This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very-large-area thin foils placed close together or using a dielectric (to be discussed below). STRATEGY FOR (B) Knowing C, find the charge stored by solving the equation, for the charge Q. Access for free at openstax.org. Solution for (b) The charge Qon the capacitor is 18.5 • Capacitors and Dielectrics 585 18.40 Discussion for (b) This charge is only slightly greater than typical static electricity charges. More charge could be stored by using a dielectric between the capacitor plates. WORKED EXAMPLE What battery is needed to charge a capacitor? Your friend provides you with a STRATEGY Use the equation capacitor. To store to find the voltage needed to charge the capacitor. on this capacitor, what voltage battery should you buy? Solution Solving for the voltage gives gives. Inserting and Discussion Such a battery should be easy to procure. There is still a question of whether the battery contains enough energy to provide the desired charge. The equation allows us to calculate the required energy. 18.42 18.41 A typical commercial battery can easily provide this much energy. Practice Problems 23. What is the voltage on a 35 μF with 25 nC of charge? a. 8.75 × 10−13 V b. 0.71 × 10−3 V 1.4 × 10−3 V c. 1.4 × 103 V d. 24. Which voltage is across a 100 μF capacitor that stores 10 J of energy? a. −4.5 × 102 V b. 4.5 × 102 V c. ±4.5 × 102 V d. ±9 × 102 V Die
lectrics Before working through some sample problems, let’s look at what happens if we put an insulating material between the plates of a capacitor that has been charged and then disconnected from the charging battery, as illustrated in Figure 18.30. Because the material is insulating, the charge cannot move through it from one plate to the other, so the charge Qon the capacitor does not change. An electric field exists between the plates of a charged capacitor, so the insulating material becomes polarized, as shown in the lower part of the figure. An electrically insulating material that becomes polarized in an electric field is called a dielectric. 586 Chapter 18 • Static Electricity Figure 18.30 shows that the negative charge in the molecules in the material shifts to the left, toward the positive charge of the capacitor. This shift is due to the electric field, which applies a force to the left on the electrons in the molecules of the dielectric. The right sides of the molecules are now missing a bit of negative charge, so their net charge is positive. Figure 18.30 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. The molecules in the dielectric are polarized by the electric field of the capacitor. All electrically insulating materials are dielectrics, but some are betterdielectrics than others. A good dielectric is one whose molecules allow their electrons to shift strongly in an electric field. In other words, an electric field pulls their electrons a fair bit away from their atom, but they do not escape completely from their atom (which is why they are insulators). Figure 18.31 shows a macroscopic view of a dielectric in a charged capacitor. Notice that the electric-field lines in the capacitor with the dielectric are spaced farther apart than the electric-field lines in the capacitor with no dielectric. This means that the electric field in the dielectric is weaker, so it stores less electrical potential energy than the electric field in the capacitor with no dielectric. Access for free at openstax.org. 18.5 • Capacitors and Dielectrics 587 Where has this energy gone? In fact, the molecules in the dielectric act like tiny springs, and the energy in the electric field goes into stretching these springs. With the electric field thus weakened, the voltage difference between the two sides of the
capacitor is smaller, so it becomes easier to put more charge on the capacitor. Placing a dielectric in a capacitor before charging it therefore allows more charge and potential energy to be stored in the capacitor. A parallel plate with a dielectric has a capacitance of 18.43 (kappa) is a dimensionless constant called the dielectric constant. Because where capacitance increases when a dielectric is placed between the capacitor plates. The dielectric constant of several materials is shown in Table 18.1. is greater than 1 for dielectrics, the Material Dielectric Constant ( ) Vacuum 1.00000 Air 1.00059 Fused quartz 3.78 Neoprene rubber 6.7 Nylon Paper 3.4 3.7 Polystyrene 2.56 Pyrex glass Silicon oil 5.6 2.5 Strontium titanate 233 Teflon Water 2.1 80 Table 18.1 Dielectric Constants for Various Materials at 20 °C 588 Chapter 18 • Static Electricity Figure 18.31 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. Because some electric-field lines terminate and start on polarization charges in the dielectric, the electric field is less strong in the capacitor. Thus, for the same charge, a capacitor stores less energy when it contains a dielectric. WORKED EXAMPLE Capacitor for Camera Flash A typical flash for a point-and-shoot camera uses a capacitor of about capacitor plates is 100 V—that is, 100 V is placed “across the capacitor,” how much energy is stored in the capacitor? (b) If the dielectric used in the capacitor were a 0.010-mm-thick sheet of nylon, what would be the surface area of the capacitor plates? STRATEGY FOR (A). (a) If the potential difference between the, we can use the equation, to find the electric potential energy Given that stored in the capacitor. and Access for free at openstax.org. Solution for (a) Inserting the given quantities into gives 18.5 • Capacitors and Dielectrics 589 18.44 Discussion for (a) This is enough energy to lift a 1-kg ball about 1 m up from the ground. The flash lasts for about 0.001 s, so the power delivered by the capacitor
during this brief time is power, this is not bad for a little capacitor!. Considering that a car engine delivers about 100 kW of STRATEGY FOR (B) Because the capacitor plates are in contact with the dielectric, we know that the spacing between the capacitor plates is. From the previous table, the dielectric constant of nylon is. We can now use the equation to find the area Aof the capacitor. Solution (b) Solving the equation for the area Aand inserting the known quantities gives 18.45 Discussion for (b) This is much too large an area to roll into a capacitor small enough to fit in a handheld camera. This is why these capacitors don’t use simple dielectrics but a more advanced technology to obtain a high capacitance. Practice Problems 25. With 12 V across a capacitor, it accepts 10 mC of charge. What is its capacitance? a. 0.83 μF b. 83 μF 120 μF c. d. 830 μF 26. A parallel-plate capacitor has an area of 10 cm2 and the plates are separated by 100 μm. If the capacitor contains paper between the plates, what is its capacitance? a. b. c. d. 3.3 × 10−10 F 3.3 × 10−8 F 3.3 × 10−6 F 3.3 × 10−4 F Check Your Understanding 27. If the area of a parallel-plate capacitor doubles, how is the capacitance affected? a. The capacitance will remain same. b. The capacitance will double. c. The capacitance will increase four times. d. The capacitance will increase eight times. 28. If you double the area of a parallel-plate capacitor and reduce the distance between the plates by a factor of four, how is the capacitance affected? 590 Chapter 18 • Static Electricity a. b. c. d. It will increase by a factor of two. It will increase by a factor of four. It will increase by a factor of six. It will increase by a factor of eight. Access for free at openstax.org. Chapter 18 • Key Terms 591 KEY TERMS capacitor arrangement of objects that can store electrical energy by virtue of their geometry conductor material through which electric charge can easily move, such as metals Coulomb’s law describes the electrostatic force between charged objects, which is proportional to the charge on each object and inversely proportional to the square of
the distance between the objects of negative electric charge induction creating an unbalanced charge distribution in an object by moving a charged object toward it (but without touching) insulator material through which a charge does not move, such as rubber inverse-square law law that has the form of a ratio, with the denominator being the distance squared dielectric electrically insulating material that becomes law of conservation of charge states that total charge is polarized in an electric field constant in any process electric field defines the force per unit charge at all locations in space around a charge distribution polarization separation of charge induced by nearby excess charge electric potential the electric potential energy per unit proton subatomic particle that carries the same magnitude charge charge as the electron, but its charge is positive electric potential energy the work that a charge can do by test charge positive electric charge whose with a charge virtue of its position in an electric field electron subatomic particle that carries one indivisible unit magnitude so small that it does not significantly perturb any nearby charge distribution SECTION SUMMARY 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge • Electric charge is a conserved quantity, which means it can be neither created nor destroyed. • Electric charge comes in two varieties, which are called positiveand negative. • Charges with the same sign repel each other. Charges with opposite signs attract each other. • Charges can move easily in conducting material. Charges cannot move easily in an insulating material. • Objects can be charged in three ways: by contact, by conduction, and by induction. • Although a polarized object may be neutral, its electrical charge is unbalanced, so one side of the object has excess negative charge and the other side has an equal magnitude of excess positive charge. 18.2 Coulomb's law • Coulomb’s law is an inverse square law and describes the electrostatic force between particles. • The electrostatic force between charged objects is proportional to the charge on each object and inversely proportional to the distance squared between the objects. If Coulomb’s law gives a negative result, the force is attractive; if the result is positive, the force is repulsive. • 18.3 Electric Field • The electric field defines the force per unit charge in the space around a charge distribution. • For a point charge or a sphere of uniform charge, the electric field is inversely proportional to the distance from the point charge or from the center of the sphere. • Electric-field lines never cross each other. • More force is applied to
a charge in a region with many electric field lines than in a region with few electric field lines. • Electric field lines start at positive charges and point away from positive charges. They end at negative charges and point toward negative charges. 18.4 Electric Potential • Electric potential energy is a concept similar to gravitational potential energy: It is the potential that charges have to do work by virtue of their positions relative to each other. • Electric potential is the electric potential energy per unit charge. • The potential is always measured between two points, where one point may be at infinity. • Positive charges move from regions of high potential to regions of low potential. • Negative charges move from regions of low potential to regions of high potential. 18.5 Capacitors and Dielectrics • The capacitance of a capacitor depends only on the geometry of the capacitor and the materials from which it is made. It does not depend on the voltage across the capacitor. • Capacitors store electrical energy in the electric field between their plates. 592 Chapter 18 • Key Equations • A dielectric material is an insulator that is polarized in • Putting a dielectric between the plates of a capacitor an electric field. increases the capacitance of the capacitor. KEY EQUATIONS 18.2 Coulomb's law Coulomb’s law 18.3 Electric Field electric field magnitude of electric field of point charge 18.4 Electric Potential change in electric potential energy for a charge that moves in a constant electric field electric potential energy of a charge a distance rfrom a point charge or sphere of uniform charge definition of electric potential change in electric potential for a charge that moves in a constant electric field electric potential of a charge a distance rfrom a point charge or sphere of uniform charge 18.5 Capacitors and Dielectrics capacitance energy stored in a capacitor capacitance of a parallel-plate capacitor CHAPTER REVIEW Concept Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 1. There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electric effects? a. Most objects are neutral. b. Most objects have positive charge only. c. Most objects have negative charge only. d. Most objects have excess protons. 2. Can an insulating material be used to charge a conductor? If so, how? If not, why not? a. No, an insulator cannot charge a conductor by induction. b. No, an insulating material cannot charge a
conductor. Access for free at openstax.org. c. Yes, an uncharged insulator can charge a conductor by induction. d. Yes, a charged insulator can charge a conductor upon contact. 3. True or false—A liquid can be an insulating material. a. b. true false 18.2 Coulomb's law 4. Two plastic spheres with uniform charge repel each other with a force of 10 N. If you remove the charge from one sphere, what will be the force between the spheres? a. The force will be 15 N. b. The force will be 10 N. c. The force will be 5 N. d. The force will be zero. 5. What creates a greater magnitude of force, two charges +qa distance r apart or two charges – qthe same distance apart? a. Two charges +qa distance raway b. Two charges −qa distance raway c. The magnitudes of forces are equal. 6. In Newton’s law of universal gravitation, the force between two masses is proportional to the product of the two masses. What plays the role of mass in Coulomb’s law? a. b. c. d. the electric charge the electric dipole the electric monopole the electric quadruple 18.3 Electric Field 7. Why can electric fields not cross each other? a. Many electric-field lines can exist at any given point in space. Chapter 18 • Chapter Review 593 b. true 10. True or false—The characteristics of an electric field make it analogous to the gravitational field near the surface of Earth. false a. true b. 11. An electron moves in an electric field. Does it move toward regions of higher potential or lower potential? Explain. a. It moves toward regions of higher potential because its charge is negative. It moves toward regions of lower potential because its charge is negative It moves toward regions of higher potential because its charge is positive. It moves toward regions of lower potential because its charge is positive. b. c. d. b. No electric-field lines can exist at any given point in 18.5 Capacitors and Dielectrics space. c. Only a single electric-field line can exist at any given point in space. d. Two electric-field lines can exist at the same point in space. 8. A constant electric field is (4.5 × 105 N/C)ŷ. In which direction is the force on a −20
nC charge placed in this field? a. The direction of the force is in the direction. direction. b. The direction of the force is in the c. The direction of the force is in the −ŷ direction. d. The direction of the force is in the +ŷ direction. 18.4 Electric Potential 9. True or false—The potential from a group of charges is the sum of the potentials from each individual charge. a. false Critical Thinking Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 15. If you dive into a pool of seawater through which an equal amount of positively and negatively charged particles is moving, will you receive an electric shock? a. Yes, because negatively charged particles are moving. b. No, because positively charged particles are moving. c. Yes, because positively and negatively charged particles are moving. 12. You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor? a. Energy stored in the capacitor will remain same. b. Energy stored in the capacitor will decrease. c. Energy stored in the capacitor will increase. d. Energy stored in the capacitor will increase first, and then it will decrease. 13. True or false— Placing a dielectric between the plates of a capacitor increases the energy of the capacitor. a. b. false true 14. True or false— The electric field in an air-filled capacitor is reduced when a dielectric is inserted between the plates. a. b. false true d. No, because equal amounts of positively and negatively charged particles are moving. 16. True or false—The high-voltage wires that you see connected to tall metal-frame towers are held aloft by insulating connectors, and these wires are wrapped in an insulating material. a. b. true false 17. By considering the molecules of an insulator, explain how an insulator can be overall neutral but carry a surface charge when polarized. a. Inside the insulator, the oppositely charged ends of 594 Chapter 18 • Chapter Review b. the molecules cancel each other. Inside the insulator, the oppositely charged ends of the molecules do not cancel each other. c. The electron distribution in all the molecules shifts in every possible direction, leaving an excess of positive charge on the opposite end of each molecule. d. The electron distribution in all the molecules shifts in a given direction, leaving an excess of negative charge on the opposite end
of each molecule. 18.2 Coulomb's law 18. In terms of Coulomb’s law, why are water molecules attracted by positive and negative charges? a. Water molecules are neutral. b. Water molecules have a third type of charge that is attracted by positive as well as negative charges. c. Water molecules are polar. d. Water molecule have either an excess of electrons or an excess of protons. 19. A negative lightning strike occurs when a negatively charged cloud discharges its excess electrons to the positively charged ground. If you observe a cloud-tocloud lightning strike, what can you say about the charge on the area of the cloud struck by lightning? a. The area of the cloud that was struck by lightning had a positive charge. b. The area of the cloud that was struck by lightning had a negative charge. repelled or attracted. b. No, because an electrically neutral body can be attracted but not repelled. c. Yes, because an electrically neutral body can be repelled or attracted. d. Yes, because an electrically neutral body can be repelled. 18.4 Electric Potential 22. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential? a. Voltage is the energy per unit mass at some point in space. b. Voltage is the energy per unit length in space. c. Voltage is the energy per unit charge at some point in space. d. Voltage is the energy per unit area in space. 23. Three parallel plates are stacked above each other, with a separation between each plate. If the potential difference between the first two plates is ΔV1 and the potential between the second two plates is ΔV2, what is the potential difference between the first and the third plates? a. ΔV3 = ΔV2 + ΔV1 b. ΔV3 = ΔV2 − ΔV1 c. ΔV3 = ΔV2 / ΔV1 d. ΔV3 = ΔV2×ΔV1 c. The area of the cloud that was struck by lightning is 18.5 Capacitors and Dielectrics neutral. d. The area of the cloud that was struck by lightning had a third type of charge. 18.3 Electric Field 20. An arbitrary electric field passes through a box-shaped volume. There are no charges in the box. If 11 electricfield lines enter the box, how many electric-field lines must exit the box
? a. nine electric field lines 10 electric field lines b. 11 electric field lines c. 12 electric field lines d. 21. In a science-fiction movie, a villain emits a radial electric field to repulse the hero. Knowing that the hero is electrically neutral, is this possible? Explain your reasoning. a. No, because an electrically neutral body cannot be 24. When you insert a dielectric into a capacitor, the energy stored in the capacitor decreases. If you take the dielectric out, the energy increases again. Where does this energy go in the former case, and where does the energy come from in the latter case? a. Energy is utilized to remove the dielectric and is released when the dielectric is introduced between the plates. b. Energy is released when the dielectric is added and is utilized when the dielectric is introduced between the plates. c. Energy is utilized to polarize the dielectric and is released when the dielectric is introduced between the plates. d. Energy is released to polarize the dielectric and is utilized when dielectric is introduced between the plates. Access for free at openstax.org. Problems 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 25. A dust particle acquires a charge of −13 nC. How many excess electrons does it carry? a. 20.8 × 10−28 electrons b. 20.8 × −19 electrons c. 8.1 × 1010 electrons d. 8.1 × 1019 electrons 26. Two identical conducting spheres are charged with a net charge of +5.0 qon the first sphere and a net charge of −8.0 qon the second sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now? a. −3.0q b. −1.5q c. +1.5q d. +3.0q 18.2 Coulomb's law 27. Two particles with equal charge experience a force of 10 nN when they are 30 cm apart. What is the magnitude of the charge on each particle? -5.8 × 10-10 C a. -3.2 × 10-10 C b. c. +3.2 × 10-10 C d. +1.4 × 10-5 C 28. Three charges are on a line. The left charge is q1 = 2.0
nC. The middle charge is q2 = 5.0 nC. The right charge is q3 = − 3.0 nC. The left and right charges are 2.0 cm from the middle charge. What is the force on the middle charge? a. −5.6 × 10−4 N to the left b. −1.12 × 10−4 N to the left c. +1.12 × 10−4 N to the right 5.6 × 10−4 N to the right d. 18.3 Electric Field 29. An electric field (15 N/C)ẑapplies a force (− 3 × 10–6 N)ẑ on a particle. What is the charge on the particle? a. −2.0 × 10–7 C b. 2.0 × 10–7 C Chapter 18 • Chapter Review 595 c. 2.0 × 10–8 C d. 2.0 × 10–9 C 30. Two uniform electric fields are superimposed. The first electric field is. The second electric. With respect to the positive field is xaxis, at which angle will a positive test charge accelerate in this combined field? a. 27° 54° b. c. 90° 108° d. 18.4 Electric Potential 31. You move a charge qfrom ri = 20 cm to rf = 40 cm from a fixed charge Q= 10 nC. What is the difference in potential for these two positions? a. −2.2 × 102 V b. −1.7 × 103 V c. −2.2 × 104 V d. −1.7 × 102 V 32. How much work is required from an outside agent to? move an electron from xi = 0 to xf = 20 cm in an electric field a. b. c. d. 1.6 × 10−15 J 1.6 × 10−16 J 1.6 × 10−20 J 1.6 × 10−18 J 18.5 Capacitors and Dielectrics 33. A 4.12 µF parallel-plate capacitor has a plate area of 2,000 cm2 and a plate separation of 10 µm. What dielectric is between the plates? a. b. 466, the dielectric is strontium c. 699, the dielectric is strontium nitrate d. 1,000, the dielectric is strontium chloride 1

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