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cup with confetti. 8 Turn on the Van de Graaff generator and let it run. 9 Dip the soap bubble dispenser into the soap and blow a stream of bubbles toward the charging sphere of the Van de Graaff generator. 10 Record your detailed observations in Table 10.1. 11 Ground the sphere with the grounding rod, and Demonstration Observation Explanation turn off the generator. Animal Fur Aluminium Pie Plates Foam-plastic Cup and Confetti Stream of Soap Bubbles Think About It Question 1. Using your knowledge of electricity, provide a possible explanation of the events that occurred during each demonstration. 1. How does the sphere of the Van de Graaff generator become charged? 2. Describe a situation during the demonstrations where the forces of interaction between the sphere of the generator and the various objects were: (a) attractive (b) repulsive 3. Why does touching the sphere with a grounding rod affect the charge on the sphere? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 10 Physics laws can explain the behaviour of electric charges. 511 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 512 10.1 Electrical Interactions Your world runs on electricity. The music you listen to, the movies you watch, the video games you play—all require electricity to run. Today, electricity is so familiar that you probably don’t even think about it when you turn on a light, pop a piece of bread into the toaster, or switch off the TV. Try to imagine a time before electricity was even named. People had noticed interesting effects in certain situations that seemed almost magical. The Greek philosopher Thales (624–546 BCE) recorded that when he rubbed amber (a hard fossilized form of tree resin), it could attract small pieces of straw or thread. This effect was called “electricity,” after the Greek word for amber, “elektron.” The ancient Greeks observed two important properties of electricity: • Charged objects could either attract or repel each other. These two types of interactions suggested that there must be two different types of charge. • Repulsion occurred when two similarly charged objects were placed near each other, and attraction occurred when two oppositely charged objects were placed near each other. These observations can be summarized as the law of charges: Like
charges repel and unlike charges attract. M I N D S O N Electrical Attraction Rub an ebonite rod with fur and hold the rod close to a fine stream of water from a faucet. Then rub a glass rod with silk and hold this rod close to a fine stream of water. Observe what happens in each case. Using your knowledge of charging objects, explain why the ebonite rod or the glass rod affects the water. There was little progress in understanding the nature of electricity until the 1600s, when the English scientist William Gilbert (1544–1603) performed extensive investigations. In De Magnete, his book on magnetism, Gilbert compared the effects of electricity and magnetism. He concluded that: 1. Objects only exhibit electrical effects when recently rubbed; magnetic objects do not need to be rubbed. 2. Electrified objects can attract small pieces of many types of objects; magnetic objects can attract only a few types of objects. 3. Electrified objects attract objects toward one central region; magnetic objects appear to have two poles. Although Gilbert was able to describe certain effects of electricity, he still did not know the origins of electric charges. In the 1700s, the American scientist and inventor Benjamin Franklin (1706–1790) attempted to prove that lightning in the sky was the same electricity as the spark observed when you reach for a metal info BIT The plastic in contact lenses contains etafilcon, which is a molecule that attracts molecules in human tears. This electrostatic attraction holds a contact lens on the eye. info BIT Gilbert was also a medical doctor and held the prestigious position of personal physician of Queen Elizabeth I of England. 512 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 513 door handle after shuffling across a carpet. He performed his famous kite experiment to explore whether lightning was a form of electricity (Figure 10.2). Luckily, he did not get killed, and he succeeded in drawing electricity from the clouds. He observed that lightning behaves the same way that electricity produced in the laboratory does. Through further investigations, he identified and named the two different types of electric charges as positive and negative charges. It soon became apparent that electricity is in all substances. This idea caught the imagination of many different people. Scientists studied electricity’s intriguing effects (Figure 10.3), and entrepreneurs exploited it. Magicians and carnivals featured the “mysterious” effects of electricity. Figure 10
.2 This figure is an artist’s representation of Benjamin Franklin’s famous kite experiment. info BIT Some historians think that Franklin may not have actually performed his kite experiment. They suspect that Franklin sent a description of this dangerous experiment to the Royal Society in London, England as a joke, because this British academy had largely ignored his earlier work. In 1753, the Royal Society awarded Franklin the prestigious Copley Medal for his electrical research. Figure 10.3 Boys were sometimes used in experiments such as this one in the early 1700s. The boy was suspended over the floor and electrostatically charged. His positive electric charge would attract pieces of paper. Studies to determine the nature of electricity continued. These studies were the beginning of the science of electrostatics, which is the study of electric charges at rest. It involves electric charges, the forces acting on them, and their behaviour in substances. electrostatics: the study of electric charges at rest The Modern Theory of Electrostatics Today’s theory of electrostatics and the nature of electric charges is based on the models of the atom that Ernest Rutherford (1871–1937) and Niels Bohr (1885–1962) proposed in the early 1900s. In their theories, an atom is composed of two types of charges: positively charged protons in a nucleus surrounded by negatively charged electrons. In nature, atoms have equal numbers of electrons and protons so that each atom is electrically neutral. Just as some materials are good thermal conductors or insulators, there are also good conductors and insulators of electric charges. Electrical conductivity depends on how tightly the electrons are bound to the nucleus of the atom. Some materials have electrons that are tightly bound to the nucleus and are not free to travel within the substance. These materials are called insulators. Materials that have electrons in the outermost regions of the atom that are free to travel are called conductors. insulator: material in which the electrons are tightly bound to the nucleus and are not free to move within the substance conductor: material in which electrons in the outermost regions of the atom are free to move Chapter 10 Physics laws can explain the behaviour of electric charges. 513 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 514 Figure 10.4 shows some examples of good conductors and insulators. Note that metals are usually good conductors. It is also interesting to note that a good conductor, such as
silver, can have a conductivity 1023 times greater than that of a good insulator, such as rubber. Figure 10.4 Relative electrical conductivity of some materials Relative magnitude of conductivity 108 107 103 109 1010 1012 1015 Material silver copper aluminium iron mercury carbon germanium silicon wood glass rubber Conductors F O I L (computer chips) Semiconductors (transistors) Insulators Semiconductors Materials that lie in the middle, between good conductors and good insulators, are called semiconductors. Because of their nature, they are good conductors in certain situations, and good insulators in other situations. Selenium, for example, is an insulator in the dark, but in the presence of light, it becomes a good conductor. Because of this property, selenium is very useful in the operation of photocopiers (Figure 10.5). original copy face down movable light lens A mirror positively charged paper A negatively charged toner brush selenium-coated drum heater assembly mirror positively charged Figure 10.5 Photocopiers use the semiconductor selenium in the copying process. 514 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 515 The selenium-coated drum in the photocopier is initially given a positive charge and kept in the dark to retain the charge. When a flash of light shines on a document to be copied, an image of the document is transferred to the drum. Where the document is light-coloured, the selenium is illuminated, causing it to be conductive. Electrons flow into the conductive portions of the selenium coating, leaving them uncharged. The page remains light-coloured or white. Where the document is darkcoloured, the selenium remains non-conductive, and the positive charge remains. Negatively charged “toner” powder is sprinkled on the drum and attaches to the positively charged portions of the drum. When a sheet of paper is passed over the drum, the toner transfers to the paper and an image of the document is created. This toner image is then fused on the paper with heat, and the copying process is complete. Silicon and germanium are also semiconductors. They become conductors when atoms such as gallium or arsenic are added to them. This process is called “doping” with impurities. The field of solid
state electronics, which includes components such as transistors, diodes, and silicon chips, is based on this type of semiconductor. Superconductors Recall from earlier science studies that resistance is a measure of how difficult it is for electrons to flow through a material. Materials with a low electrical resistance are better conductors because very little energy is lost to heat in the conduction of electricity. Early attempts at conducting electricity efficiently used conducting materials with low electrical resistance, such as silver, copper, and gold. Researchers soon discovered that the electrical resistance of any material tends to decrease as its temperature decreases. Could the temperature of a material be lowered to the point that it loses all its resistive nature, creating the ideal conductor? This property of materials would have an enormous range of applications. Once a current is established in such a conductor, it should persist indefinitely with no energy loss. In the early 20th century, a class of materials called superconductors was developed. These conductors have no measurable resistance at very low temperatures. The Dutch physicist Heike Kammerlingh Onnes (1853–1926) discovered this effect in 1911 when he observed that solid mercury lost its electrical resistance when cooled to a temperature of 269 C. Although this discovery was significant, the usefulness of superconductors was limited because of the extremely low temperatures necessary for their operation. It was not until 1986 that materials were developed that were superconductors at much higher temperatures. These materials are ceramic alloys of rare earth elements, such as lanthanum and yttrium. As an example, one such alloy was made by grinding yttrium, barium, and copper oxide into a mixture and heating the mixture to form the alloy YBa2Cu3O7. This substance became a superconductor at 216 C. In 1987, another alloy was developed that displayed superconductivity at 175 C. More recent discoveries have reported copper oxide alloys that are superconductors at temperatures as high as 123 C. The ultimate goal is to develop superconductors that operate at room temperature, thus creating a whole new era of useful applications in technology. e WEB Find out what research is being done on superconductors today. How soon will you be seeing superconductors in use around the house? Write a brief summary of what you discover. To learn more about superconductors, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 515
10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 516 10-2 Inquiry Lab 10-2 Inquiry Lab Charging Objects Question How can objects become electrically charged? Materials and Equipment 2 white plastic polyethylene strips (or ebonite rods) fur (approximately 15 cm x 15 cm) 2 clear plastic acetate strips (or glass rods) silk (approximately 15 cm x 15 cm) electroscope silk thread 2 retort stands with clamps Procedure Part A: Charging by Friction 1 Copy Table 10.2 into your notebook. Table 10.2 Observations for Charging by Friction White Polyethylene Strip Rubbed with Fur Clear Acetate Strip Rubbed with Silk Hanging White Polyethylene Strip Hanging Clear Acetate Strip Hanging White Polyethylene Strip Hanging Clear Acetate Strip 2 Hang a white polyethylene strip from one retort stand and a clear acetate strip from another retort stand. 3 While holding the hanging white polyethylene strip in the middle, rub both ends of it with the fur. While holding the hanging clear acetate strip in the middle, rub both ends of it with the silk. 4 Rub the other white polyethylene strip with the fur. 5 Carefully bring this second polyethylene strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. 6 Carefully bring the second polyethylene strip close to one end of the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. 7 Observe what happens in each situation and record your observations in Table 10.2. 516 Unit VI Forces and Fields Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 8 Rub a clear acetate strip with the silk. 9 Carefully bring this strip close to the hanging clear acetate strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. 10 Carefully bring this clear acetate strip close to one end of the hanging white polyethylene strip. Do not allow the two plastic strips to touch each other. Observe what happens and record your observations in Table 10.2. Part B: Charging by Conduction 11 Copy Table 10.3 into your notebook. Table 10.3 Observations for Charging by Conduction Electroscope Charged with the White
Polyethylene Strip Rubbed with Fur Electroscope Charged with the Clear Acetate Strip Rubbed with Silk 12 Rub the unattached white polyethylene strip with the fur. Touch this white strip to the knob of the electroscope. 13 Carefully bring the electroscope close to one end of the hanging white polyethylene strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 14 Now bring the electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves in the electroscope. Record your observations in Table 10.3. 15 Ground the electroscope by touching the knob of the electroscope with your finger. 16 Rub a clear acetate strip with the silk and touch the strip to the knob of the grounded electroscope. 17 Repeat steps 13 and 14. 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 517 Part C: Charging by Induction 18 Copy Table 10.4 into your notebook. Table 10.4 Observations for Charging by Induction Grounded Electroscope Hanging White Polyethylene Strip Hanging Clear Acetate Strip 19 Bring an uncharged electroscope near one end of the hanging white polyethylene strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. 20 Bring an uncharged electroscope near one end of the hanging clear acetate strip. Observe what happens to the leaves of the electroscope, and record your observations in Table 10.4. Analysis 1. What effect did you observe when two similarly charged white polyethylene strips were held near each other or when two similarly charged clear acetate strips were held near each other? 2. What effect did you observe when a charged white polyethylene strip was held near an oppositely charged hanging clear acetate strip or when a charged clear acetate strip was held near an oppositely charged hanging white polyethylene strip? 3. Based on your observations, what charge did the electroscope acquire when it was touched by the charged white polyethylene strip? when it was touched by the charged acetate strip? 4. What evidence shows a movement of charge in the electroscope when it is held near a charged object? 5. From your observations in Table 10.2, what general rule can you formulate about attraction and repulsion of charged objects? 6. From your observations in Table 10.3, what general
rule can you formulate about the charge received by an object when it is touched by another charged object? 7. From your observations in Table 10.4, what general rule can you formulate about the charge received by an object when it is held near another charged object? 8. Does the electroscope acquire a net electrical charge during the process of charging by induction? Justify your answer. 9. What evidence is there from this investigation to prove that there are two types of electrical charges? 10. From the investigation, is there any evidence to prove which type of charge was developed on the white polyethylene strip and on the clear acetate strip? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Methods of Charging Objects According to the modern theory of electrostatics, objects can become charged through a transfer of electrons. Electron transfer can occur in three ways: by friction, by conduction, and by induction. Law of Conservation of Charge During any charging procedure, it is important to keep in mind that new charges are not being created. The charges existing in materials are merely being rearranged between the materials, as the law of conservation of charge states: The net charge of an isolated system is conserved. Net charge is the sum of all electric charge in the system. For example, if a system contains 3 C of charge and 5 C of charge, the system’s net coulomb (C): SI unit for charge, equivalent to the charge on 6.25 1018 electrons or protons Chapter 10 Physics laws can explain the behaviour of electric charges. 517 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 518 charge is 2 C. Suppose you have a system that initially consists of two electrically neutral objects, and there is a transfer of electrons from one object to the other. One object will lose electrons and become positively charged while the other object will gain these electrons and become equally negatively charged. However, the net charge of the system is still zero. Charges have not been created, they have only been rearranged. Charging Objects by Friction The most common method of charging objects is by rubbing or friction. You have probably had the unpleasant experience of receiving a shock when you touched a door handle after walking across a carpeted floor. Similarly, gently stroking a cat can result in the generation of small sparks, which are very uncomfortable for the cat. Charging by this
method involves separating electrons from the atoms in one object through rubbing or friction, and then transferring and depositing these electrons to the atoms of another object. The object whose atoms lose electrons then possesses positively charged ions. The object whose atoms gain electrons possesses negatively charged ions. As shown in Figure 10.6, rubbing the ebonite rod with fur transferred some of the electrons in the fur to the rod. The fur becomes positively charged, and the rod becomes negatively charged. (a) (b) hold electrons tightly sulfur brass copper ebonite paraffin wax silk lead fur wool glass Figure 10.6 (a) A neutral ebonite rod and a neutral piece of fur have equal amounts of negative and positive charge. When the fur is rubbed against the rod, a transfer of electrons occurs. (b) After rubbing, the ebonite has gained electrons and has a net negative charge. The fur has lost electrons and has a net positive charge. Whether an object gains or loses electrons when rubbed by another object depends on how tightly the material holds onto its electrons. Figure 10.7 shows the electrostatic series, in which substances are listed according to how tightly they hold their electrons. Substances at the top have a strong hold on their electrons and do not lose electrons easily. Substances near the bottom have a weak hold on their electrons and lose them easily. hold electrons loosely Concept Check Figure 10.7 The electrostatic or triboelectric series 1. Using information from Figure 10.7, explain why ebonite acquires a greater charge when rubbed with fur rather than silk. 2. What type of charge does ebonite acquire when rubbed with fur? Charging objects by friction can also occur during collisions. The collisions of water vapour molecules in rain clouds, for example, cause 518 Unit VI Forces and Fields conduction: process of charging an object through the direct transfer of electrons when a charged object touches a neutral object 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 519 the separation and transfer of electrons. The result is that vapour molecules become positively or negatively charged, eventually resulting in lightning. You will learn more about lightning later in this chapter. This process of charging objects was also observed by the Voyageur spacecraft on its mission to Saturn. Colliding particles in the rings of Saturn create electrical discharges within the rings, similar to lightning on Earth. Charging Objects by Conduction Objects can become charged by the transfer of electrons
from a charged object to an uncharged object by simply touching the objects together (Figure 10.8). This process is called charging by conduction. (a) (b) Figure 10.8 (a) During charging by conduction, electrons from a negatively charged metal conducting sphere transfer to a neutral metal conducting sphere, upon contact. (b) The neutral sphere gains electrons and is said to have been charged by conduction. The quantity of charge that transfers from one object to another depends on the size and shape of the two objects. If both objects are roughly the same size and shape, the charge transferred will be such that both objects are approximately equally charged (Figure 10.9(a)). If one sphere is larger than the other, then the larger sphere will receive more of the charge (Figure 10.9(b)). When the spheres are separated, the excess charges move to become equidistant from each other because of the forces of repulsion between like charges. Charging by conduction is similar to charging by friction because there is contact between two objects and some electrons transfer from one object to the other. (a) (b) Figure 10.9 Electrostatic repulsion of like charges forces excess charges within objects to redistribute so that the distances between charges are equal. (a) If two objects are the same size, the charges redistribute equally. (b) If the two objects are different sizes, the object with a larger surface area has more charges. Once the charge has transferred to another object, it will either be distributed over the surface of the object, if the object is a conductor, or remain on the surface at the point of contact, if the object is an insulator. Chapter 10 Physics laws can explain the behaviour of electric charges. 519 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 520 excess negative charge neutral Figure 10.10 A piece of paper appears to be attracted to a charged ebonite rod, even before they touch. induction: movement of charge caused by an external charged object Charging Objects by Induction If you bring a negatively charged ebonite rod slowly toward a small piece of uncharged paper, the rod will attract the piece of paper, as shown in Figure 10.10. In fact, the piece of paper would begin to jiggle and move toward the rod even before the rod touches it. This reaction is a result of the forces acting on electrostatic charges. You know that electrostatic attraction can occur only
between oppositely charged objects, but how can a charged object attract a neutral or uncharged object? And why is there never a force of repulsion between a charged object and a neutral object? The answers to these questions are revealed in the third method of charging objects, which involves two processes: induction and charging by induction. Induction Induction is a process in which charges in a neutral object shift or migrate because of the presence of an external charged object. This temporary charge separation polarizes the neutral object. One side of the object becomes positively charged and the other side is equally negatively charged. Although the object now behaves as if it is charged, it is still electrically neutral. The charging object and the neutral object do not touch each other, so there is no actual transfer of charge. (a) (b) negatively charged negatively charged Figure 10.11 (a) A neutral metal sphere and a neutral piece of paper (b) The influence of the large negative charge of a rod causes charge migration within the conducting sphere, which polarizes the sphere. The influence of the rod causes charge shift within the atoms of the insulating paper. The atoms in the paper become polarized. Because of induction, the sides of the sphere and the atoms in the paper that are positively charged are closer to the negatively charged rod than their negatively charged sides are. The net result is attraction. The process of induction varies slightly, depending on whether the charging object is approaching a substance that is an insulator or a conductor. Figure 10.11(a) depicts a neutral metal sphere (conductor) and a neutral piece of paper (insulator). Figure 10.11(b) shows a negatively charged rod approaching each neutral object. The electrons in the two neutral objects are repelled by the negative charge of an ebonite rod. The metal sphere is a conductor, so the electrons can move easily through it to its other side. This process of charge migration causes the sphere to become polarized, where one side of the sphere is positive and the other side is negative. charge migration: movement of electrons in a neutral object where one side of the object becomes positive and the other side becomes negative 520 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 521 Since the paper is an insulator, its electrons cannot move easily through it, so they just shift slightly relative to the nuclei. This process of charge shift causes the atoms to become polarized, where one side of an atom
becomes positive and the other side becomes negative. In both cases, the distance from the negatively charged rod to the positive end of the neutral object is less than the distance to the negative end of the object. Therefore, the attraction of the opposite charges is greater than repulsion of like charges, and the net force is attractive. Charge separation by induction, which results in polarization of objects, explains electrostatic situations such as the initial attraction of a neutral piece of paper to a negatively charged rod without contact, as you saw in Figure 10.10. Charging by Induction In the situation shown in Figure 10.11, the electrons in the metal sphere and the paper return to their original positions when the negatively charged rod is removed. The objects lose their polarity and remain electrically neutral. For conductors, like the metal sphere, it is possible to maintain a residual charge by adding a grounding step. Grounding involves touching or connecting a wire from the object to the ground, as shown in Figure 10.12(a). The grounding path is then removed while the source charge is still present. The grounding step allows the conductor to maintain a charge (Figure 10.12(b)). The complete process of charging by induction includes grounding. (a) (b) charge shift: movement of electrons in an atom where one side of an atom becomes positive and the other side becomes negative grounding: the process of transferring charge to and from Earth charging by induction: the process of polarizing an object by induction while grounding it Figure 10.12 (a) While the charged rod is held near the metal sphere, the sphere remains polarized by induction. Grounding the sphere removes excess charge. In this situation, the sphere appears to have excess electrons on one side, which are removed. The positive charges cannot move because they represent the fixed nuclei of atoms. (b) After the ground and charged rod are removed, the sphere retains a net positive charge because of the loss of electrons. It has been charged by induction. PHYSICS INSIGHT The symbol for ground is A grounded conductor that is polarized by the presence of a charged object will always have a net charge that is opposite to that of the charged object if the ground is removed before the charged object is removed. The conductor has been charged by induction. Chapter 10 Physics laws can explain the behaviour of electric charges. 521 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 522 Concept Check When you rub a balloon on your hair, you are
charging it by friction. Explain why the balloon will then stick to a wall for a long time. How Lightning Gets Its Charge Many theories attempt to explain the formation of lightning. One theory relates the cause to the processes of evaporation and condensation of water in the clouds and different methods of charging objects. Under the right conditions, a churning cloud formation causes water vapour molecules to collide, resulting in a transfer of electrons between these molecules. The transfer of an electron from one water molecule to another leaves the molecules oppositely charged. plasma: highly ionized gas containing nearly equal numbers of free electrons and positive ions Cooling causes water vapour molecules to condense into water droplets. The atoms in these droplets hold onto electrons more readily than atoms in water vapour, and thus the droplets become negatively charged. Being heavier, these negatively charged water droplets accumulate at the bottom of the cloud, causing the bottom of the cloud to become negatively charged (Figure 10.13). The top of the cloud, containing the rising water vapour, becomes positively charged. The increasing polarization of the cloud ionizes the surrounding air, forming a conductive plasma. Excess electrons on the bottom of the cloud begin a zigzag journey through this plasma toward the ground at speeds of up to 120 km/s, creating a step leader. This is not the actual lightning strike. The presence of the large negative charge at the bottom of the cloud causes the separation of charges at that location on Earth’s surface. Earth’s surface at that spot becomes positively charged, and the area below the surface becomes negatively charged. Charge separation has polarized Earth’s surface. Air molecules near Earth’s surface become ionized and begin to drift upward. This rising positive charge is called a streamer. When the rising positive streamer meets the step leader from the clouds, at an altitude of about 100 m, a complete pathway is formed and the lightning begins. A transfer of negative charge in the form of a lightning strike from the cloud travels to Earth’s surface at speeds of up to 100 000 km/s (Figure 10.14). Figure 10.13 Lightning forms when the bottom of the cloud becomes negatively charged and Earth’s surface becomes positively charged. Figure 10.14 A streamer moving up from Earth’s surface meets a step leader coming down from the clouds and lightning lights up the sky. 522 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24
/08 2:51 PM Page 523 10.1 Check and Reflect 10.1 Check and Reflect Knowledge 1. What is the science of electrostatics? 2. Describe a simple experiment that enabled early scientists to determine that there were two different types of charges. 3. In the 1600s, William Gilbert compared the effects of electricity and magnetism. (a) Describe two similarities between 7. Describe how you would charge the sphere in Figure 10.12 negatively by induction. 8. A negatively charged ebonite rod is brought near a neutral pith ball that is hanging by an insulating thread from a support. Describe what happens (a) before they touch (b) after they touch these effects. 9. Compare the distribution of charge (b) Describe two differences between these effects. 4. In the classification of substances by electrical conductivity, a substance may be a conductor, insulator, semiconductor, or superconductor. (a) What property of matter determines the electrical conductivity of a substance? (b) List the classifications given above in order of increasing electrical conductivity. (c) Give an example of a substance in each classification. (d) Describe the conditions when the semiconductor selenium becomes a conductor or an insulator. (a) on hanging aluminium and glass rods if both are touched at one end by a negatively charged ebonite rod (b) after a small negatively charged metal sphere momentarily touches a larger neutral metal sphere 10. A negatively charged ebonite rod is held near the knob of a neutral electroscope. (a) Explain what happens to the leaves of the electroscope. (b) Explain what happens to the leaves of the electroscope if the other side of the knob is now grounded while the rod is still in place. (c) Explain why removing the ground first and then the rod will leave a net charge on the electroscope. Applications Extensions 5. (a) An ebonite rod is rubbed with fur. 11. You are given an ebonite rod, fur, an How can the electrostatic series chart in Figure 10.7 on page 518 help you determine which object will become negatively charged? (b) Why is it better to rub an ebonite rod with fur rather than silk? 6. Describe how you could charge a glass sphere positively using the following methods: electroscope, and a sphere of unknown charge. Describe an experimental procedure that you could use to determine the charge on
the sphere. 12. If a glass rod becomes positively charged when rubbed with silk, use the law of conservation of charge to explain why the silk must be negatively charged. (a) friction (b) conduction (c) induction e TEST To check your understanding of electrical interactions, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 523 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 524 info BIT Charles de Coulomb was given credit for investigations into the electrostatic forces acting on charged objects, but the actual discovery of the relationship between electrostatic forces and charged objects was made earlier by Henry Cavendish (1731–1810), an English scientist. However, he was so shy that he didn’t publish his results and thus was not credited with the discovery. PHYSICS INSIGHT Vertical bars around a vector symbol are an alternative method for representing the magnitude of a vector. This notation is used for the magnitude of force and field vectors from this chapter on. This notation avoids confusing the symbol for energy, E, with the magnitude of the electric field strength, which is introvector, E duced in the next chapter. 10.2 Coulomb’s Law In Chapter 4, you studied Newton’s law of universal gravitation and learned that any two objects in the universe exert a gravitational force g). The magnitude of this force of gravitational attraction on each other (F is directly proportional to the product of the two masses (m1 and m2): F g m1m2 and inversely proportional to the square of the distance between their centres (r): F g 1 r 2 These relationships can be summarized in the following equation: F g G m1m2 r 2 where G is the universal gravitational constant in newton-metres squared per kilogram squared. Charles de Coulomb suspected that the gravitational force that one mass exerts on another is similar to the electrostatic force that one charge exerts on another. To verify his hypothesis, he constructed an apparatus called a torsion balance to measure the forces of electrostatics. Although it could not be used to determine the quantity of charge on an object, Coulomb devised an ingenious method to vary the quantity of charge in a systematic manner. 10-3 Inquiry Lab 10-3 Inquiry Lab Investigating the Variables in Coulomb’s Law Required Skills Initi
ating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Two charged objects exert electrostatic forces of magnitude Fe on each other. How does Fe depend on the charges carried by the objects and on the separation between the objects? Materials and Equipment Van de Graaff generator or ebonite rod and fur 3 small Styrofoam™ or pith spheres, about 1 cm in diameter, coated with aluminium or graphite paint Hypothesis State a hypothesis relating the electrostatic force and each of the variables. Remember to write an “if/then” statement. Variables Read the procedure for each part of the inquiry and identify the manipulated variable, the responding variable, and the controlled variables in each one. sewing needle about 65 cm of thread 3 drinking straws 2 retort stands and 2 clamps balance 2 rulers tape small mirror (about 10 cm long and 5 cm wide) marking pen 524 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 525 Procedure 1 Determine the mass of one sphere (sphere 1) and record this mass in kilograms in your notebook. 2 Using the sewing needle, draw the thread through the centre of this sphere and attach both ends of the thread, with tape, to both ends of a drinking straw so that the sphere is suspended in the centre with the thread forming a V pattern. Clamp this straw, horizontally, to a retort stand, as shown in Figure 10.15. 3 Carefully insert the second drinking straw into the second sphere (sphere 2). Then fasten this drinking straw, horizontally, to the clamp on the other retort stand. 7 Tape the mirror on the paper so that the image of sphere 1 aligns with the centre of the mirror. Tape a ruler over the mirror just below the image of sphere 1. Using the marking pen, mark the position of the centre of the image of sphere 1 on the mirror while sighting in such a way that the sphere lines up with its image. 8 Do part A of the activity, which begins below, followed by part B. mirror line indicating top position of thread attached to straw image of sphere 1 string sphere 1 sphere 3 4 Adjust the clamps so that both spheres are at the paper same height. sphere 2 5 Carefully insert the third drinking straw into the third sphere (sphere 3). (This sphere will be the grounding sphere that will be used to change the charges on spheres 1 and 2.)
6 Tape sheets of white paper to the wall. Place the retort stand with sphere 1 close to the wall, about 5 cm away, so that the centre of the sphere aligns with the centre of the paper. Using a ruler, draw a horizontal line on the paper indicating the top position of the string attached to the straw. Figure 10.15 Part A: The relationship between the quantity of charge on each object and the electrostatic force In this part of the lab, the distance between the spheres is held constant while the charges on the spheres are varied. 9 Copy Table 10.5 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.5 Data and Calculations for Part A Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Product of Charges of Spheres 1 and 2 (q2) Magnitude of Force of Electrostatic Repulsion Acting on Spheres 1 and 2 (N) e F (0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0250) (0.0130) (0.0062) (0.0030) q 1 q1 1 2 2 4 q 1 q1 1 4 8 2 1 q1 1 q 6 1 4 4 q1 1 1 q 8 4 2 3 1 q1 1 q 4 6 8 8 Chapter 10 Physics laws can explain the behaviour of electric charges. 525 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 526 10 Rub the ebonite rod with the fur or charge the Van de Graaff generator. 11 Carefully touch sphere 2 to the charged rod or generator to charge the sphere by conduction. Since it is nearly impossible to measure the quantity of charge transferred to sphere 2, assume that the quantity of charge on the sphere is q. 15 To obtain more data, vary the charge on each sphere using sphere 3. Gently touch sphere 3 to sphere 1. Since the charge on each sphere should
be shared equally, the new charge on sphere 1 is 1 q. The 4 q 1 q 1 charge product on spheres 1 and 2 is now 1 q 2. 8 2 4 Label the new position of sphere 1 as position 2. 12 Slide the stand holding charged sphere 2 toward 16 Remove sphere 3 to a safe distance and ground it by sphere 1 on the other stand and momentarily touch the two spheres together. The charge on each object can be assumed to be 1 q because, on contact, the charge 2 is equally divided between two similar spheres. q 1 q 1 Therefore, the charge product is 1 q 2. 4 2 2 13 Slide sphere 2, parallel to the wall, to the left until the centre of its image is 1.0 cm away from the centre mark on the mirror. 14 Mark the new position of the centre of the image of sphere 1 on the mirror. Label it position 1. gently touching it with your hand. 17 Repeat steps 14 to 16, keeping sphere 2 in position and alternately touching spheres 1 and 2 with sphere 3 to obtain three more readings. Label these positions 3, 4, and 5. 18 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the top horizontal line. Record this vertical distance (dy ) in metres in the appropriate column of Table 10.5. 19 Measure the distance between the original centre mark on the mirror and the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.5. Part B: The relationship between the distance between two charges and the electrostatic force In this part of the lab, the charges on the spheres are fixed while the distances are varied. 20 Copy Table 10.6 into your notebook, leaving out the numbers in parentheses in the first three columns. Use these numbers only if you are unable to measure the distances accurately. They are hypothetical values that you can use to complete the rest of the activity. Table 10.6 Data and Calculations for Part B Magnitude of Weight of Sphere 1 F g (103 N) F mg g (1.28) (1.28) (1.28) (1.28) (1.28) Vertical Height of Sphere 1 dy (m) Horizontal Distance to Centre of Sphere 1 from the Centre Mark dx (m) Distance Between the Centres of Spheres 1 and 2 r (m) Magnitude of Force of Electrostatic Repulsion Acting on
Spheres 1 and 2 (N) e F (0.300) (0.300) (0.300) (0.300) (0.300) (0.0500) (0.0467) (0.0438) (0.0409) (0.0383) (0.0600) (0.0617) (0.0638) (0.0659) (0.0683) 526 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 527 21 Rub the ebonite rod with the fur or charge the Van de 2. Construct a graph of the force of electrostatic Graaff generator. 22 Carefully touch sphere 2 to the rod or the generator to give it a charge. 23 Slide the stand holding charged sphere 2 toward sphere 1 on the other stand and momentarily touch the two spheres together. repulsion on the y-axis as a function of the charge product on the x-axis. 3. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the charge product? 4. Complete the calculations as indicated in Table 10.6 for 24 Slide sphere 2 to a position so that the centre of its Part B. image is 0.5 cm to the left of the original centre mark on the mirror. 25 Mark the new position of the image of sphere 1 on the mirror and label it position 1. 26 Repeat steps 24 and 25, sliding sphere 2 parallel to the wall so that its image positions are at 1.0 cm, 1.5 cm, 2.0 cm, and 2.5 cm away, to obtain four more readings. Label these positions 2, 3, 4, and 5. 27 With a ruler, accurately measure the vertical distance from the centre of the image of sphere 1 to the horizontal line representing the top position of the string. Record this vertical distance (dy) in metres in the appropriate column of Table 10.6. 28 Measure the distance from the original centre mark on the mirror to the centre of the image of sphere 1 in its new position for each trial. Record this distance (dx) in metres in the appropriate column of Table 10.6. 29 Measure the distance between the centres of the two spheres. For each trial, record this distance (r) in metres in the appropriate column of Table 10.6. Analysis 1. Although the force of electrostatic repulsion
F e acting on two similarly charged spheres cannot be measured directly, it can be calculated as shown in Figure 10.16. Complete the calculations as indicated in Table 10.5 for Part A. 5. Construct a graph of the force of electrostatic repulsion on the y-axis as a function of the distance between the charges on the x-axis. 6. What does the shape of the graph indicate about the relationship between the force of electrostatic repulsion and the distance between the two charges? 7. From the investigation, identify the two variables that affect the force of electrostatic repulsion acting on two charges. 8. Using a variation statement, describe the relationship between these two variables and the force of electrostatic repulsion. 9. Does your investigation confirm your hypotheses about the relationship between the variables and the electrostatic force? Why or why not? 10. How does the relationship between the variables affecting the electrostatic force in this investigation compare with that of the variables affecting the force of gravitational attraction in Newton’s law of gravitation? 11. (a) What sources of error could have led to inaccuracy in the investigation? (b) What modifications to the investigation would you recommend? dy dx Fg dy Fe dx Fe Fg Figure 10.16(a) Use the concept of similar triangles: F dy dx dx dy e F g F e F g Figure 10.16(b) The shaded triangles are similar. Chapter 10 Physics laws can explain the behaviour of electric charges. 527 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 528 The Force of Electrostatic Repulsion or Attraction Coulomb correctly hypothesized that the two factors influencing the magnitude of the electrostatic force that one charge exerts on another were the magnitudes of the charges on each object and their separation distance. To experimentally derive the relationships between the two factors and the electrostatic force, Coulomb used a procedure similar to that used in the 10-3 Inquiry Lab but with a different apparatus—the torsion balance shown in Figure 10.17. To determine the force of electrostatic attraction or repulsion acting on two charged objects, a charged ball on a rod is brought near a charged object at one end of the arm of the torsion balance. Repulsion or attraction causes the ball on the arm to move, rotating the arm. As the arm rotates, a sensitive spring either tightens or loosens, causing the needle to move a proportional angle.
This movement of the needle can be measured on a scale. The amount of movement is related to a measure of the force of electrical attraction or repulsion. Figure 10.17 Coulomb’s apparatus Determining Relative Charge e MATH In 10-3 Inquiry Lab, Investigating the Variables in Coulomb’s Law, you learned how separation and the magnitude of electric charges affect the electrostatic force. To graph the electrostatic force as a function of separation, and to analyze this relationship in more depth, visit www.pearsoned.ca/school/ physicssource. e SIM Explore the inverse square relationship through a simulation using a sphere of uniform charge density. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Realizing that there was not yet any way of measuring charge, Coulomb devised a method of accurately determining the relative magnitude of a charge. He knew that if a charged object with a charge of q touches a similar uncharged object, then the charge would be shared equally so that each object would have a charge of 1 q. Using this 2 assumption, he was able to do his experiment. Investigating the relationship between the electrostatic force and the distance between the centres of the spheres, he first charged a sphere with a charge q and touched it momentarily to the sphere on the torsion balance. Each sphere would then have a similar and equal q and 1 charge of 1 q. Then, holding the first sphere a measured distance 2 2 from the sphere in the torsion balance, he was able to measure the electrostatic force acting on the two spheres by the movement of the needle on the calibrated scale. Changing the distance between the spheres and measuring the force each time, he demonstrated there was an inverse square relationship between the electrostatic force and the separation distance. This relationship can be expressed as F e 1 r 2 Investigating further the relationship between the magnitude of the force and the magnitudes of the charges, he was able to accurately vary the charges on each sphere. By charging one sphere with a charge q and touching it to the sphere on the balance, he knew that the charge would be shared equally. The two spheres would have charges of 1 q and 1 q 2 2 q)(1 each, and the charge product would be (1 q). By touching each 2 2 charged sphere alternately with a third neutral and similar sphere, he q)(1 q), then (1 q)(1 could vary the charge products as (1
q), and so on. 4 4 2 4 528 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 529 By varying the charges on both objects and measuring the electrostatic force acting on them, he demonstrated that the magnitude of the electrostatic force is proportional to the product of the two charges: F e q1q2 In 1785, using the results from his experimentation on charged objects, Charles de Coulomb summarized his conclusions about the electrostatic force. This force is also known as the Coulomb force. His summary of his conclusions is called Coulomb’s law. The magnitude of the force of electrostatic attraction or repulsion (F e • directly proportional to the product of the two charges q1 and q2: ) is: F e q1q2 • inversely proportional to the square of the distance between their centres r : F e 1 2 r If these are the only variables that determine the electrostatic force, then F e q1q2 r 2 The beautiful fact about Coulomb’s law and Newton’s law of gravitation is that they have exactly the same form even though they arise from different sets of operations and apply to completely different kinds of phenomena. The fact that they match so exactly is a fascinating aspect of nature. Although Coulomb was able to identify and determine the relationships of the variables that affect the electrostatic force acting on two charges, he was unable to calculate the actual force. To do so, the variation statement must be converted to an equation by determining a proportionality constant (k), whose value depends on the units of the charge, the distance, and the force. At the time, however, it was impossible to measure the exact quantity of charge on an object. The Magnitude of Charges The SI unit for electric charge is the coulomb (C). A bolt of lightning might transfer 1 C of charge to the ground, while rubbing an ebonite rod with fur typically separates a few microcoulombs (C). It is difficult to build up larger quantities of charge on small objects because of the tremendous repulsive forces between the like charges. As you will see in section 15.2, experiments at the beginning of the 20th century showed that an electron has a charge of about 1.60 1019 C. So, 1 C of negative charge corresponds to the charge on 6.25 1018 electrons, or 6.25 billion billion electrons. Similarly, the charge
on a proton is about 1.60 1019 C. PHYSICS INSIGHT Newton’s law of gravitation is called an inverse square law because the gravitational force acting on any two masses is inversely proportional to the square of the distance between their centres. Chapter 10 Physics laws can explain the behaviour of electric charges. 529 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 530 info BIT The electrostatic force acting on two charges of 1 C each separated by 1 m is about 9 109 N. This electrostatic force is equal to the gravitational force that Earth exerts on a billion-kilogram object at sea level! Given these values, physicists were able to calculate the constant of proportionality for Coulomb’s law. With this constant, Coulomb’s law becomes: F e 1q q 2 k 2 r is the magnitude of the force of electrostatic attraction or where F e repulsion in newtons; q1 and q2 are the magnitudes of the two charges in coulombs; r is the distance between the centres of the charges in metres; k is the proportionality constant called Coulomb’s constant and is equal to 8.99 109 Nm2/C2. This electrostatic force is attractive if the two objects have opposite charges and repulsive if the two objects have like charges. This equation is used to determine electrostatic forces in many different types of problems involving charges and the electrostatic forces acting on them. Examples 10.1 and 10.2 show how to calculate the electrostatic force of attraction or repulsion acting on two charges in a one-dimensional situation. Example 10.1 Practice Problem 1. In a hydrogen atom, an electron is 5.29 1011 m from a proton. An electron has a charge of 1.60 1019 C, and the proton’s charge is 1.60 1019 C. Calculate the electrostatic force of attraction acting on the two charges. Answer 1. 8.22 108 N [attraction] A small metal sphere with a negative charge of 2.10 106 C is brought near an identical sphere with a positive charge of 1.50 106 C so that the distance between the centres of the two spheres is 3.30 cm (Figure 10.18). Calculate the magnitude and type (attraction or repulsion) of the force of one charge acting on another. Given q1 2.10 106 C 1.50 106
C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Required magnitude and type of the electrostatic force acting on the two charges (F e ) Figure 10.18 Analysis and Solution According to Newton’s third law, the electrostatic forces acting on the two spheres are the same in magnitude but opposite in direction. The magnitude of the electrostatic force is F e 1q q 2 k 2 r m (2.10 106 C)(1.50 106 C) 8.99 109 N C (3.30 102 m)2 2 2 26.0 N The magnitude calculation does not use the positive and negative signs for the charges. However, you can use these signs to determine whether the electrostatic force is attractive or repulsive. In this example, the charges have opposite signs, so the force is attractive. 530 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 531 Paraphrase The electrostatic force is one of attraction, with a magnitude of 26.0 N. In the next example, the two spheres touch and the charge is distributed between them. Example 10.2 The two spheres in Example 10.1 are momentarily brought together and then returned to their original separation distance. Determine the electrostatic force now exerted by one charge on the other. Given initial magnitude of the charges: 2.10 106 C q1 1.50 106 C q2 r 3.30 102 m 3.30 cm 2.10 106 C 1.50 106 C Figure 10.19 Practice Problem 1. A metal sphere with a negative charge of 3.00 C is placed 12.0 cm from another similar metal sphere with a positive charge of 2.00 C. The two spheres momentarily touch, then return to their original positions. Calculate the electrostatic force acting on the two metal spheres. Answer 1. 1.56 101 N [repulsion] Required magnitude and type of the electrostatic force acting on the e) two charges (F Analysis and Solution When a sphere with a negative charge of 2.10 106 C momentarily touches a sphere with a positive charge of 1.50 106 C, then 1.50 106 C of charge from the first sphere neutralizes the 1.50 106 C of charge on the second sphere. The remaining charge of 0.60 106 C from the first sphere then divides equally between the two identical spheres. Each sphere now has a charge of 3.
0 107 C. The magnitude of the electrostatic force is now F e 1q q 2 k 2 r m (3.0 107 C)(3.0 107 C) 8.99 109 N 2 C (3.30 102 m)2 2 0.74 N Since both spheres have a negative charge, the electrostatic force is repulsive. Paraphrase The electrostatic force is one of repulsion, with a magnitude of 0.74 N. Chapter 10 Physics laws can explain the behaviour of electric charges. 531 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 532 Concept Check Compare gravitational forces and electrostatic forces by identifying two similarities and two differences between the two types of forces. Vector Analysis of Electrostatic Forces So far in this section, you have studied Coulomb’s law and applied the equation to calculate the magnitude of the electrostatic force that one charged particle exerts on another. However, many situations involve more than two charges. The rest of this section illustrates how to use Coulomb’s law to analyze the vector nature of electrostatic forces by determining the electrostatic forces of more than two charges in onedimensional and two-dimensional situations. Examples 10.3 and 10.4 illustrate how to apply Coulomb’s law to three or more collinear charges. Recall from unit I that collinear entities lie along the same straight line. Example 10.3 A small metal sphere (B) with a negative charge of 2.0 106 C is placed midway between two similar spheres (A and C) with positive charges of 1.5 106 C that are 3.0 cm apart (Figure 10.20). Use a vector diagram to find the net electrostatic force acting on sphere B. Analysis and Solution 1.5 cm 1.5 cm A 1.5 106 C B 2.0 106 C C 1.5 106 C Figure 10.20 Spheres A and C have equal charges and are the same distance from sphere B. As shown in Figure 10.21, the force vectors are equal in length and opposite in direction. FAB B FCB Figure 10.21 F net F F CB F AB CB F AB, so F net 0. Since the forces are equal in magnitude and opposite in direction, the net electrostatic force on charge B is 0. Practice Problems 1. Three small, hollow, metallic spheres hang from insulated threads as shown in the figure below. Draw a free-body diagram
showing the electrostatic forces acting on sphere B. 1.0 cm 2.0 cm A 2.0 μC B 2.0 μC C 2.0 μC 2. For the figure in problem 1 above, draw a vector for the net electrostatic force on sphere B. Answers 1. FAB Fnet 2. B FCB 532 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 533 Example 10.4 A small metal sphere (B) with a negative charge of 2.10 106 C is placed midway between two similar spheres (A and C) 3.30 cm apart with positive charges of 1.00 106 C and 1.50 106 C, respectively, as shown in Figure 10.22. If the three charges are along the same line, calculate the net electrostatic force on the negative charge. Given qA qB qC rAC rAB 1.00 106 C 2.10 106 C 1.50 106 C 3.30 102 m 1 rAC rBC 2 3.30 cm 1.65 cm 1.00 106 C 2.10 106 C 1.50 106 C Figure 10.22 Required net) net electrostatic force on qB (F Analysis and Solution The charge on sphere B is negative and the charge on sphere A is positive, so the electrostatic force of qA on qB, AB, is an attractive force to the left. Similarly, the F CB, is an attractive force electrostatic force of qC on qB, F to the right (Figure 10.23). Consider right to be positive. FAB B FCB Figure 10.23 F The sum of these two force vectors is the net force on qB: F F net Applying F CB 1q q 2 gives k 2 r AB e F net 2 m (1.00 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 Practice Problems 1. A metal sphere with a charge of 2.50 109 C is 1.50 cm to the left of a second metal sphere with a charge of 1.50 109 C. A third metal sphere of 1.00 109 C is situated 2.00 cm to the right of the second charged sphere. If all three charges form a line, determine the net electrostatic force on the second sphere. 2. In the situation described above, if the first and third spheres remain at their
original positions, where should the second sphere be situated so that the net electrostatic force on it would be zero? Answers 1. 1.16 104 N [to the left] 2. 2.14 102 m to the right of the 2.50 109 C charge [left] 2 m (1.50 106 C)(2.10 106 C) 8.99 109 N C 3.30 102 m2 2 2 [right] (69.34 N 104.0 N) [right] 34.7 N [right] Paraphrase The net electrostatic force on charge B is 34.7 N to the right. In Examples 10.3 and 10.4, the forces act along the same line, so the calculations involve only a single dimension. Examples 10.5 and 10.6 demonstrate how to calculate net electrostatic forces in two dimensions. Chapter 10 Physics laws can explain the behaviour of electric charges. 533 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 534 Example 10.5 Practice Problems 1. A small metal sphere X with a negative charge of 2.50 C is 1.20 cm directly to the left of another similar sphere Y with a charge of 3.00 C. A third sphere Z with a charge of 4.00 C is 1.20 cm directly below sphere Y. The three spheres are at the vertices of a right triangle, with sphere Y at the right angle. Calculate the net electrostatic force on sphere Y, sketching diagrams as necessary. 2. Calculate the net electrostatic force on charge B shown in the figure below. 2.00 μC A 1.00 cm 90° 1.00 μC B Answers 1. 8.83 1014 N [122] 2. 2.54 102 N [225] 1.00 cm C 2.00 μC A small metal sphere A with a negative charge of 2.10 106 C is 2.00 102 m to the left of another similar sphere B with a positive charge of 1.50 106 C. A third sphere C with a positive charge of 1.80 106 C is situated 3.00 102 m directly below sphere B (Figure 10.24). Calculate the net electrostatic force on sphere B. 2.00 102 m A 2.10 106 C B 1.50 106 C Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C
2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere B (F Analysis and Solution The electrostatic force of qA on qB, F AB, is a force of attraction directed from charge B toward charge A (left). The electrostatic CB, is a force force of qC on qB, F of repulsion directly upward (Figure 10.25). 3.00 102 m C 1.80 106 C Figure 10.24 FCB B FAB Figure 10.25 Applying F e 1q q 2 gives k 2 r F AB 2.10 106 C)(1.50 106 C) 8.99 109 N C (2.00 102 m)2 2 2 70.80 N Similarly, F CB 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 534 Unit VI Forces and Fields 26.97 N 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 535 Use trigonometry to find the net electrostatic force on charge B, as shown in Figure 10.26. Use the Pythagorean theorem to find the magnitude of the net force: F net (70.80 N)2 (26.97 N)2 75.8 N Determine the angle : 9 N 7. 6 tan 2 7 N 0 8. 0 20.9 Fnet FCB θ FAB Figure 10.26 The direction of the net force is [20.9 N of W] or [159]. Paraphrase The net electrostatic force on charge B is 75.8 N [20.9 N of W], or 75.8 N [159]. PHYSICS INSIGHT Recall that with the polar coordinates method, angles are measured counterclockwise from the positive x-axis of the coordinate system, which is given a value of 0. Practice Problems 1. Three metal spheres are situated in positions forming an equilateral triangle with sides of 1.20 cm, as shown below. X has a charge of 2.50 C; Y has a charge of 3.00 C; and Z has a charge of 4.00 C. Calculate the net electrostatic force on the Y charge. X 2.50 C 1.20 cm 1.20 cm Y 3.00 C 1.20 cm Z 4.00 C 2. Four charged spheres, with equal charges of 2.20 C, are
situated in positions forming a rectangle, as shown in the figure below. Determine the net electrostatic force on the charge in the top right corner of the rectangle. 2.00 102 m A 2.10 106 C B 1.50 106 C rAC 3.00 102 m C 1.80 106 C Figure 10.27 Example 10.6 A small metal sphere A with a charge of 2.10 106 C is 2.00 102 m to the left of a second sphere B with a charge of 1.50 106 C. A third sphere C with a charge of 1.80 106 C is situated 3.00 102 m directly below sphere B. Calculate the net electrostatic force on sphere C. Given qA qB qC rAB rBC 2.10 106 C 1.50 106 C 1.80 106 C 2.00 102 m 3.00 102 m Required net electrostatic force net) on sphere C (F FAC Analysis and Solution The electrostatic force of qA on qC, AC, is an attractive force directed F from charge C toward charge A. The electrostatic force of qB on qC, BC, is a repulsive force directed F downward (Figure 10.28). C 2.20 C 40.0 cm 2.20 C 30.0 cm 30.0 cm 40.0 cm 2.20 C 2.20 C FBC Figure 10.28 Answers 1. 6.56 1014 N [142°] 2. 7.17 1011 N [55.0°] Chapter 10 Physics laws can explain the behaviour of electric charges. 535 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 536 Determine the distance between charges A and C by using the Pythagorean theorem (Figure 10.29): rAC (2.00 102 m)2 (3.00 102 m)2 3.606 102 m 3.61 102 m Applying F e 1q q 2 gives k 2 r 2.00 102 m B A 3.00 102 m θ 1 C F AC 2.10 106 C)(1.80 106 C) 8.99 109 N C (3.61 102 m)2 Figure 10.29 2 2 26.13 N Similarly, F BC 1.50 106 C)(1.80 106 C) 8.99 109 N C (3.00 102 m)2 2 2 26.97 N Use the component method to find the
sum of the two force vectors. Use trigonometry to determine the angle 1 AC (Figure 10.29): for the direction of F tan 1 1 2 0 0. 3 0 0. 33.69 2 m 1 2 1 m 0 0 Then resolve F AC into x and y components, as shown in Figure 10.30: FACx FACy (26.13 N) (sin 33.69) 14.49 N (26.13 N) (cos 33.69) 21.74 N x 26.13 N y 33.7° C Figure 10.30 The electrostatic force of charge B on charge C has only a y component (see Figure 10.28). So, the x component of F BC is 0 N and the y component is 26.97 N. net. Now find the sum of the x and y components of F F net Fnetx F F BC AC FACx FBCx 14.49 N 0 N 14.49 N Fnety FBCy FACy 21.74 N (26.97 N) 5.23 N 536 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 537 Use trigonometry to determine the magnitude and direction of the net electrostatic force on charge C, as shown in Figure 10.31. Determine the magnitude of the net force using the Pythagorean theorem: (14.49 N)2 (5.23 N)2 15.4 N F net 5.23 N 14.49 N θ 2 Fnet C Figure 10.31 To determine the angle 2, use the tangent function: tan 2 2 3 N.2 5 1 N 9.4 4 19.8 The direction of the net force is [19.8 S of W] or [200]. Paraphrase The net electrostatic force on charge C is 15.4 N [19.8 S of W] or 15.4 N [200]. THEN, NOW, AND FUTURE ESD Control Manager Since the early 1970s, electrostatic discharge (ESD) has evolved from an interesting, but generally harmless, phenomenon to one of the most rapidly expanding fields of research in science today. As electronic devices have become smaller and smaller, ESD has become a major cause of failure. Each year, billions of dollars’ worth of electronic devices and systems are destroyed or degraded by electrical stress caused by ESD. A dangerous property of E
SD is its ability to cause fires in a flammable atmosphere. Property loss, injuries, and fatalities due to the accidental ignition of petrochemical vapours, dusts, and fuels by ESD are on the rise. ESD has been the proven ignition source in many fires. However, research into the firesparking nature of ESD is still in its infancy. Today, electronics manufacturers have ESD awareness and control programs, ESD control program managers, and, in some cases, entire departments dedicated to preventing the damaging effects of ESD. Ron Zezulka (Figure 10.32) is the chief technical officer of TB&S Consultants and has specialized in the science of ESD for over 30 years. He graduated from the Southern Alberta Institute of Technology with a diploma as a telecommunications technician and began his career as a failure analyst specializing in the for Alberta science of ESD Government Telephones. Ron completed many courses to become a control program manager. Because of the newness of the industry, there are no specific quali- Figure 10.32 Ron Zezulka fications for becoming a control program manager in the field of ESD. An ESD control program manager might have a technical diploma and related job experience, a Master’s degree, or a Ph.D. in physics. In 2001, Ron formed TB&S Consultants and has developed and delivered over 25 different training programs and management systems for the awareness and control of ESD in industry. He has written on the topic and lectured in industry, universities, and colleges on awareness and control of ESD. Static electricity is now tied to almost every aspect of the physical sciences. As technology advances, so does our need for a greater understanding of ESD phenomena. Questions 1. Describe two hazards associated with ESD. 2. How could ESD have damaging and harmful effects in your home? 3. How are ESD control program managers employed in industry? Chapter 10 Physics laws can explain the behaviour of electric charges. 537 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 538 10.2 Check and Reflect 10.2 Check and Reflect Knowledge 1. Identify the two factors that influence the force of electrostatic attraction or repulsion acting on two charges. Write a mathematical expression to describe the relationship. 2. Describe how the inverse square law, first proposed by Newton for gravitational forces, was applied to electrostatic forces by Coulomb. 3. (a)
What is the SI unit for electric charge? (b) Compare the charge on an electron to that produced by rubbing an ebonite rod with fur. 4. Coulomb could not measure the amount of charge on his spheres, but he could vary the amount of charge on each sphere. Describe the procedure he used to do so. Applications 5. An electrostatic force of 10 N acts on two charged spheres, separated by a certain distance. What will be the new force in the following situations? 7. Two identical conducting spheres have charges of 5.00 105 C and 6.00 105 C and are in fixed positions, 2.00 m apart. (a) Calculate the electrostatic force acting on the two charges. (b) The spheres are touched together and returned to their original positions. Calculate the new electrostatic force acting on them. 8. Three charges are placed in a line, as shown in the diagram below. 2.00 m 3.00 m A 2.00 mC B 3.00 mC C 2.00 mC (a) What is the net electrostatic force on charge A? (b) What is the net electrostatic force on charge B? Extensions (a) The charge on one sphere is doubled. 9. A helium nucleus has a positive charge (b) The charge on both spheres is doubled. 6. (a) Why is it difficult to attain a large charge of 100 C on a small object? (b) During the rubbing process, an object acquires a charge of 5.0 109 C. How many electrons did the object gain? with a magnitude twice that of the negative charge on an electron. Is the electrostatic force of attraction on an electron in a helium atom equal to the force acting on the nucleus? Justify your answer. 10. Electrical forces are so strong that the combined electrostatic forces of attraction acting on all the negative electrons and positive protons in your body could crush you to a thickness thinner than a piece of paper. Why don’t you compress? e TEST To check your understanding of Coulomb’s law, follow the eTest links at www.pearsoned.ca/school/physicssource. 538 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 539 CHAPTER 10 SUMMARY Key Terms and Concepts law of conservation of charge net charge conduction induction charge migration charge shift grounding charging by induction plasma
Coulomb’s law coulomb (C) electrostatics insulator conductor semiconductor superconductor Key Equation F e 1q q 2 k 2 r Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. types of materials conductors insulators Electrostatics forces between charges Coulomb’s law calculating Fe between 2 charges methods of charging objects calculating Fe between more than 2 charges in one dimension friction induction charge separation grounding Chapter 10 Physics laws can explain the behaviour of electric charges. 539 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 540 CHAPTER 10 REVIEW Knowledge 1. (10.1) What is an electrostatic charge? 2. (10.1) On what property of materials does thermal and electrical conductivity depend? 3. (10.1) Under what conditions does selenium become a good conductor or a good insulator? What are materials with this property called? 4. (10.1) What are three methods of charging objects? 5. (10.1) During the process of charging objects by friction, what determines which object becomes negatively or positively charged? 6. (10.1) How are the processes of charging objects by conduction and friction alike? How are they different? 7. (10.1) State the law of charges. 8. (10.1) Who is credited with first naming the two types of charges as negative and positive charges? 9. (10.1) State the law of conservation of charge. 10. (10.1) Selenium and germanium are both semiconductors. Explain why selenium is used in photocopiers rather than germanium. 11. (10.2) Calculate the electrostatic force acting on two charged spheres of 3.00 C and 2.50 C if they are separated by a distance of 0.200 m. Applications 12. What is the distance between two charges of 5.00 C each if the force of electrostatic repulsion acting on them is 5.00 103 N? 13. Charge A has a charge of 2.50 C and is 1.50 m to the left of charge B, which has a charge of 3.20 C. Charge B is 1.70 m to the left of a third charge C, which has a charge of 1.60 C. If all three
charges are collinear, what is the net electrostatic force on each of the following? (a) charge B (b) charge C 14. Why is dust attracted to the front of a cathode- ray tube computer monitor? 15. Why is it desirable to develop materials with low electrical resistance? 16. Explain why a charged ebonite rod can be discharged by passing a flame over its surface. 17. Explain why repulsion between two objects is the only evidence that both objects are charged. 18. Why do experiments on electrostatics not work well on humid days? 19. Why does a charged pith ball initially attract a neutral pith ball, then repel it after touching it? 20. Why can you not charge a copper rod while holding one end with one hand and rubbing the other end with a piece of fur? 21. A person standing on an insulated chair touches a charged sphere. Is the person able to discharge the sphere and effectively ground it? Explain. 22. Two charged spheres, separated by a certain distance, attract each other with an electrostatic force of 10 N. What will be the new force in each of the following situations? (a) The charge on both spheres is doubled and the separation distance is halved. (b) The charge on one sphere is doubled while the charge on the other sphere is tripled and the separation distance is tripled. 23. Calculate and compare the electrical and gravitational forces acting on an electron and a proton in the hydrogen atom when the distance between their centres is 5.29 1011 m. 24. An equilateral triangle with sides of 0.200 m has three charges of 2.50 C each, situated on the vertices of the triangle. Calculate the net electrostatic force on each charge. What assumption did you have to make to complete the calculation? 540 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 541 25. In a Coulomb-type experiment, students were investigating the relationship between the force of electrostatic repulsion acting on two charged spheres and their separation distance. The results of their investigation yielded the results shown in the table below. Separation Distance (r) ( 102 m) Magnitude of Force of Repulsion F (N) 1.00 2.00 3.00 4.00 5.00 360.0 89.9 40.0 27.5 14.4 (a) Draw a graph of the results shown in the
table. (b) From the shape of the graph, what is the relationship between the electrostatic force and the separation distance between two charges? (c) Make a new table of values to obtain data to straighten the graph. (d) Draw a graph of the data in your new table of values. (e) Determine the slope of the graph. (f) What value does the slope of this graph represent? (g) If the charges of the two spheres are the same, what is the value of the charge on each sphere? Extensions 26. Can a neutral object contain any charges? Explain. 27. Is it possible for a single negative or a single positive charge to exist in nature under normal conditions? Explain your answer. 28. Explain why it is impossible to charge a coin by rubbing it between your fingers. 29. Compare the production of lightning on Earth with the lightning between the rings of Saturn observed by the Voyager spacecraft on its mission to Saturn. 30. You are given two equally sized metal spheres on insulated stands, a piece of wire, a glass rod, and some silk. Devise and describe a method to do the following without touching the rod to the spheres: (a) give the spheres equal and opposite charges (b) give the spheres equal and like charges 31. Using the principles of electrostatics, explain the causes and effects of the following demonstrations: (a) Two strips of clear adhesive tape are stuck together and then carefully separated. When the two strips are brought close to each other, attraction occurs. (b) Two strips of clear adhesive tape are stuck onto a desktop and then carefully removed. When the two strips are held close to each other, repulsion occurs. Consolidate Your Understanding Create your own summary of the behaviour of electric charges and the laws that govern electrical interactions by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 539 and the Learning Outcomes on page 510. 1. Create a flowchart to describe how to calculate the electrostatic forces between two or more charged objects in one- or two-dimensional situations. 2. Write a paragraph explaining the three methods of charging objects. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 511. How would you answer each question now? e TEST To check your understanding of the behaviour of electric charges, follow the eTest links at www.pe
arsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 541 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 542 C H A P T E R 11 Key Concepts In this chapter, you will learn about: vector fields electric fields electric potential difference moving charges in electric fields Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define vector fields compare forces and fields compare, qualitatively, gravitational and electric potential energy define electric potential difference as a change in electric potential energy per unit of charge calculate the electric potential difference between two points in a uniform electric field explain, quantitatively, electric fields in terms of intensity (strength) and direction relative to the source of the field and to the effect on an electric charge describe, quantitatively, the motion of an electric charge in a uniform electric field explain electrical interactions, quantitatively, using the law of conservation of charge Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and new technologies may lead to scientific discovery 542 Unit VI Electric field theory describes electrical phenomena. Figure 11.1 The eerie glow of St. Elmo’s fire on the masts of a ship On Christopher Columbus’s second voyage to the Americas, his ships headed into stormy weather, and the tips of the ships’ masts began to glow with a ghostly bluish flame. Sailors of the time believed that this bluish glow was a good sign that the ship was under the protection of St. Elmo, the patron saint of sailors, so they called the blue “flames” St. Elmo’s fire (Figure 11.1). People throughout history have written about this strange glow. Julius Caesar reported that “in the month of February, about the second watch of the night, there suddenly arose a thick cloud followed by a shower of hail, and the same night the points of the spears belonging to the Fifth Legion seemed to take fire.” Astronauts have seen similar glows on spacecraft. What is the cause of this eerie phenomenon? Why does it most often appear during thunderstorms? You will discover the answers to these questions as you continue to study the phenomena associated with electric charges. In this chapter, you will begin by learning how knowledge of the forces related to electric charges led to the
idea of fields, and you will compare different types of electric fields. Then you will learn how force is used to define the strength of electric fields. Finally, you will study the motion of charges in electric fields and explain electrical interactions using the law of conservation of energy. 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 543 11-1 QuickLab 11-1 QuickLab Shielding of Cellular Phones Electronic equipment usually contains material that is used as “shielding.” In this activity, you will discover what this shielding material does. 3 Remove the aluminium foil and again dial the number of the cellular phone. 4 Repeat steps 1 to 3 using the sheets of other materials. Problem How does the shielding of electronic equipment, such as a cellular phone, affect its operation? Materials 2 cellular phones sheets (about 20 cm × 20 cm) of various materials, such as aluminium foil, plastic wrap, wax paper, paper, cloth, fur 1 short length of coaxial cable Procedure Part A 1 Wrap the sheet of aluminium foil around one of the cellular phones. 2 With the other cellular phone, dial the number of the wrapped cellular phone and record any response. Part B 5 Carefully remove the outer strip of insulated plastic around one end of the coaxial cable and examine the inner coaxial cable wires. Questions 1. What effect did wrapping a cellular phone with the various materials have on the operation of the cellular phone? 2. Cellular phones receive communication transmissions that are electrical in nature. Speculate why the transmissions are shielded by certain materials. Which materials are most effective for shielding? 3. What material forms the protective wrapping around the inner coaxial transmission wires? Explain the purpose of this protective wrapping. Think About It 1. Desktop computers or computers in vehicles have sensitive electronic components that must be protected from outside electrical interference. Identify a possible source of outside electrical interference. Describe how computer components may be protected from this interference and explain why this protection is necessary. 2. Sometimes, if your debit card fails to scan, the clerk wraps the card with a plastic bag and re-scans it. Explain why a plastic bag wrapped around a card would allow the card to scan properly. Why do clerks not wrap the card with aluminium foil for re-scanning? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 11 Electric field
theory describes electrical phenomena. 543 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 544 11.1 Forces and Fields Figure 11.2 Forces exerted by the horses attached to the chariot cause the “violent” motion of the chariot. The ancient Greek philosophers explained most types of motion as being the result of either “violent” or “natural” forces. They thought that violent forces cause motion as the result of a force exerted by one object in contact with another (Figure 11.2). They thought that natural forces cause the motion of objects toward their “natural element” (Figure 11.3). However, the Greeks found another kind of motion more difficult to explain. You will observe this kind of motion in the following Minds On activity. M I N D S O N Action at a Distance Charge a rubber rod by rubbing it with fur and slowly bring it close to the hairs on your forearm. Do not touch the hairs or your arm. Observe what happens. 1. What evidence is there that the charged rod affects the hairs on your arm without actual contact? Is the force exerted by the rod on the hairs of your arm attractive or repulsive? 2. The rubber rod seems to be able to exert a type of violent force on the hairs of your arm without visible contact. This type of force was classified as “action at a distance,” where one object could exert a force on another object without contact. To explain “action at a distance,” the Greeks proposed the effluvium theory. According to this theory, all objects are surrounded by an effluvium. This invisible substance is made up of minute string-like atoms emitted by the object that pulsate back and forth. As the effluvium extends out to other bodies, the atoms of the different objects become entangled. Their effluvium eventually draws them toward each other. The effluvium theory helped to explain what seemed to be “action at a distance.” Although the effluvium was invisible, there was still a form of contact between the objects. Figure 11.3 To return to its natural element, a rock falls with “natural” motion to Earth’s surface. info BIT A new theory in physics, called string theory, proposes that objects interact through “strings” that transmit the forces between the objects. This new theory has a striking similarity to the
effluvium theory proposed 2500 years ago. 544 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 545 Fields In the 17th century, scientists, including Newton, tried to determine why one object can exert a force on another object without touching it. These scientists attempted to explain “action at a distance,” such as the curved path of a thrown ball or the effect of a charged piece of amber on the hair on a person’s arm. Finding that “natural” or “violent” forces and “effluvium” could not explain gravity or electrical forces, scientists developed the concept of fields to describe these forces. A field is defined as a region of influence surrounding an object. The concept of fields helps explain the laws of universal gravitation, which you studied in Chapter 4. Consider a space module on its way to the Moon (Figure 11.4). Nearing its lunar destination, the module begins to experience the increasing influence of the Moon. As a result, the module’s motion begins to follow a curved path, similar to the projectile motion of an object thrown horizontally through the air near Earth’s surface. As Newton’s laws state, the motion of any object can follow a curved path only when acted on by a non-zero force that has a perpendicular component. In space, this happens to the space module when it is near the Moon, so the space near the Moon must be different from the space where no large objects like the Moon are present. From this, we can infer that a field exists around a large object, such as the Moon. When other objects enter this field, they interact with the Moon. Similarly, Earth has a field. Gravitational force acts on other objects that enter this field. Recall from Chapter 4 that this field around objects is called a gravitational field. field: a region of influence surrounding an object Michael Faraday (1791–1867) developed the concept of fields to explain electrostatic phenomena. He determined that the space around a rubber rod must be different when the rubber rod is charged than when it is not. The charges on the rod create an electric field around the rod. An electrostatic force acts on another charged object when it is placed in this field. An electric field exists around every charge or charged object. It can exist in empty space, whether or not another charge or charged object is in the field.
Although field theory is a powerful tool for describing phenomena and predicting forces, physicists are still debating how objects can actually exert forces at a distance. Chapter 17 describes how quantum theory provides an extremely accurate model for describing such forces. Figure 11.4 A space module passing near a large planet or the Moon follows a curved path. Concept Check Use field theory to explain the path of a baseball thrown from outfield to home plate. Chapter 11 Electric field theory describes electrical phenomena. 545 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 546 11-2 Inquiry Lab 11-2 Inquiry Lab Electric Field Patterns — Demonstration Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the shape of the electric field around various charged objects? Materials and Equipment plastic platform with 2 electrode holders overhead projector petri dish canola or olive oil lawn seeds single-point electrode two-point (oppositely charged) electrodes parallel copper plates about 4 cm × 4 cm hollow sphere conductor 4–6 cm in diameter 2 Wimshurst generators connecting wires Procedure 1 Pour some of the canola or olive oil into the petri dish so the dish is about three-quarters full. 2 Place the petri dish with the oil on the plastic platform on the overhead projector. Carefully sprinkle the lawn seeds evenly over the surface of the oil. 3 Attach the single-point electrode, with a connecting wire, to one contact of the Wimshurst generator. Immerse the electrode in the oil in the centre of the dish. 4 Crank the Wimshurst generator several times and carefully observe the pattern of the seeds in the oil. 5 Remove the electrode and allow sufficient time for the lawn seeds to redistribute on the surface. (Gentle stirring with a pencil might be required.) 6 Repeat steps 3 to 5 with each of the following: (a) (b) (c) two electrodes connected to similar contacts on two Wimshurst machines two electrodes connected to opposite contacts on one Wimshurst machine two parallel copper plates connected to opposite contacts on one Wimshurst machine (d) one hollow sphere connected to one contact of one Wimshurst machine Analysis 1. Describe and analyze the pattern of the lawn seeds created by each of the charged objects immersed in the oil in step 6 of the procedure by answering the following questions: (a) Where does the density of the lawn seeds appear to be the greatest? the least? (b) Does
there appear to be a starting point and an endpoint in the pattern created by the lawn seeds? 2. Are there any situations where there appears to be no observable effect on the lawn seeds? 3. Based on your observations of the patterns created by the lawn seeds on the surface of the oil, what conclusion can you make about the space around charged objects? Magnitude and Direction of an Electric Field The electric field that surrounds a charged object has both magnitude and direction. Therefore, an electric field is classified as a vector field. At any point around a charge, the field can be represented by a vector arrow. The arrow’s length represents the magnitude of the electric field and the arrowhead indicates direction at that point. By definition, the direction of the electric field around a charge is the direction of the force experienced by a small positive test charge placed in the electric field (Figure 11.5). A test charge is a charge with a magnitude small enough so that it does not disturb the charge on the source charge and thus change its electric field. test charge: charge with a magnitude small enough that it does not disturb the charge on the source charge and thus change its electric field source charge: charge that produces an electric field 546 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 547 (a) (b) test charge F source charge source charge F test charge Figure 11.5 The direction of the electric field at a point is the direction of the electric force exerted on a positive test charge at that point. (a) If the source charge is negative, the field is directed toward the source. (b) If the source charge is positive, the field is directed away from the source. Concept Check Identify the difference in the electric field strength, E II, as represented by the vector arrows in Figure 11.6., at points I and II E I E Figure 11.6 You can determine the magnitude of the electric field around a point charge from the effect on another charge placed in the field. If a small positive test charge is placed in the field, this charge will experience a greater force when it is near the charge producing the field than when it is farther away from it. By definition, the electric field (E ) at a given point is the ratio of the electric force (F e) exerted on a charge (q) placed at that point to the magnitude of that charge. The electric field can be calculated using the equation
F e E q where q is the magnitude of the test charge in coulombs (C); F e is the electric force on the charge in newtons (N); and E is the strength of the electric field at that point in newtons per coulomb (N/C), in the direction as defined previously. info BIT A tremendous range of field strengths occurs in nature. For example, the electric field 30 cm away from a light bulb is roughly 5 N/C, whereas the electron in a hydrogen atom experiences an electric field in the order of 1011 N/C from the atom’s nucleus. Chapter 11 Electric field theory describes electrical phenomena. 547 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 548 Example 11.1 A sphere with a negative charge of 2.10 × 10–6 C experiences an electrostatic force of repulsion of 5.60 × 10–2 N when it is placed in the electric field produced by a source charge (Figure 11.7). Determine the magnitude of the electric field the source charge produces at the sphere. Practice Problems 1. An ion with a charge of 1.60 10–19 C is placed in an electric field produced by another larger charge. If the magnitude of the field at this position is 1.00 103 N/C, calculate the magnitude of the electrostatic force on the ion. 2. The magnitude of the electrostatic force on a small charged sphere is 3.42 10–18 N when the sphere is at a position where the magnitude of the electric field due to another larger charge is 5.34 N/C. What is the magnitude of the charge on the small charged sphere? Answers 1. 1.60 10–16 N 2. 6.40 10–19 C 5.60 102 N F 2.10 106 C Figure 11.7 source charge Given q 2.10 106 C F e 5.60 102 N [repulsion] Required magnitude of the electric field (E ) Analysis and Solution F e, q Since.67 104 N/C Paraphrase The magnitude of the electric field is 2.67 104 N/C at the given point. q1 q2 r Figure 11.8 A test charge (q2) is placed in the electric field of a source charge (q1). The distance between their centres is r. PHYSICS INSIGHT Equations based on Coulomb’s law only work for point charges. 548
Unit VI Forces and Fields The equation for determining the magnitude of the electric field around a point charge, like that shown in Figure 11.8, can be derived mathematically as follows: F e and F e q 2 kq 1 2 r q2, then If E q2 kq 1 2 r E k E r q 2 where q is the magnitude of the source charge producing the electric field in coulombs (ignore the sign of the charge); r is the distance from the 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 549 centre of the source charge to a specific point in space in metres; k is Coulomb’s constant (8.99 109 Nm2/C2); and E is the magnitude of the electric field in newtons per coulomb. Example 11.2 Determine the electric field at a position P that is 2.20 10 –2 m from the centre of a negative point charge of 1.70 10 –6 C. Given q 1.70 106 C r 2.20 102 m Required electric field E Analysis and Solution The source charge producing the electric field is q. So, q k E 2 r (1.70 106 C) 8.99 109 N m 2 2 C (2.20 102 m)2 3.16 107 N/C Practice Problems 1. The electric field at a position 2.00 cm from a charge is 40.0 N/C directed away from the charge. Determine the charge producing the electric field. 2. An electron has a charge of 1.60 10–19 C. At what distance from the electron would the magnitude of the electric field be 5.14 1011 N/C? Answers 1. 1.78 1012 C 2. 5.29 1011 m Since the source charge is negative and the field direction is defined as the direction of the electrostatic force acting on a positive test charge, the electric field is directed toward the source charge. Paraphrase The electric field at point P is 3.16 107 N/C [toward the source]. Concept Check Compare gravitational fields and electrostatic fields by listing two similarities and two differences between the two types of fields. Often, more than one charge creates an electric field at a particular point in space. In earlier studies, you learned the superposition principle for vectors. According to the superposition principle, fields set up by many sources superpose to form a single net field. The vector
specifying the net field at any point is simply the vector sum of the fields of all the individual sources, as shown in the following examples. Example 11.3 shows how to calculate the net electric field at a point in one-dimensional situations. e MATH The nucleus of an atom exhibits both electric and gravitational fields. To study their similarities and differences graphically, visit www.pearsoned.ca/school/ physicssource. Chapter 11 Electric field theory describes electrical phenomena. 549 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 550 Example 11.3 Practice Problems 1. Calculate the net electric field at a point 2.10 10–2 m to the left of the 1.50 10–6 C charge in Figure 11.9. 2. An electron and a proton are 5.29 10–11 m apart in a hydrogen atom. Determine the net electric field at a point midway between the two charges. Answers 1. 3.67 107 N/C [left] 2. 4.11 1012 N/C [toward the electron] Two positively charged spheres, A and B, with charges of 1.50 10 –6 C and 2.00 10 –6 C, respectively, are 3.30 10 –2 m apart. Determine the net electric field at a point P located midway between the centres of the two spheres (Figure 11.9). 1.50 106 C 2.00 106 C A P 3.30 102 m B Figure 11.9 Given qA qB r 3.30 102 m 1.50 106 C 2.00 106 C Required net electric field at point P (E net) Analysis and Solution As shown in Figure 11.10, the electric field created by qA at point P is directed to the right, while the electric field at point P created by qB is directed to the left. Consider right to be positive. The distance between qA and point P is: rqA to P 3.30 102 m 1.65 102 m 2 To calculate the electric field at point P created by qA, use: (1.50 106 C) 8.99 109 m 2 2 C N (1.65 102 m)2 E qA q A k r 2 q to P A 4.953 107 N/C To calculate the electric field at point P created by qB, use: (2.00 106 C) 8.99 109 m 2 2
C N (1.65 102 m)2 E qB q B k r 2 q to P B 6.604 107 N/C Use vector addition to determine the net electric field at point P: E net E qA E qB 4.953 107 N/C [right] 6.604 107 N/C [left] 1.65 107 N/C [left] Paraphrase The net electric field at point P is 1.65 107 N/C [left]. P Eq B Eq A Figure 11.10 550 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 551 Example 11.4 demonstrates how to determine the net electric field at a point due to two charges in a two-dimensional situation. Example 11.4 Calculate the net electric field at a point P that is 4.00 10–2 m from a small metal sphere A with a negative charge of 2.10 10–6 C and 3.00 10–2 m from another similar sphere B with a positive charge of 1.50 10–6 C (Figure 11.11). P Figure 11.11 4.00 102 m 3.00 102 m A 36.9° 2.10 106 C 53.1° B 1.50 106 C Given qA rA to P A 2.10 106 C 4.00 102 m 36.9 to the horizontal qB rB to P B 1.50 106 C 3.00 102 m 53.1 to the horizontal Required net) net electric field at point P (E Analysis and Solution Since qA is a negative charge, the electric field created by qA at point P is directed toward qA from point P. Since qB is a positive charge, the electric field created by qB at point P is directed away from qB toward point P. Determine the electric field created by qA at point P: E A q k A 2 A r to P 2 m (2.10 106 C) 8.99 109 N 2 C (4.00 102 m)2 1.180 107 N/C Determine the electric field created by qB at point P: E B q k B 2 B r to P 2 m (1.50 106 C) 8.99 109 N C (3.00 102 m)2 2 Practice Problems 1. Calculate the net electric field at point P, which is 0.100 m from
two similar spheres with positive charges of 2.00 C and separated by a distance of 0.0600 m, as shown in the figure below. P 0.100 m 0.100 m 72.5° 72.5° 2.00 C 0.0600 m 2.00 C 2. Two charges of +4.00 C are placed at the vertices of an equilateral triangle with sides of 2.00 cm, as shown in the figure below. Determine the net electric field at the third vertex of the triangle. 2.00 cm 2.00 cm 60° 60° 4.00 C 2.00 cm 4.00 C Answers 1. 3.43 1012 N/C [90.0°] 2. 1.56 1014 N/C [90.0°] 1.498 107 N/C Chapter 11 Electric field theory describes electrical phenomena. 551 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 552 B are shown in Figure 11.12. A and E The directions of E y EB 53.1° 36.9° EA x 36.9° y P x EAy EAx EA y EB EBy 53.1° x P EBx Figure 11.12 Figure 11.13 Figure 11.14 Resolve each electric field into x and y components (see Figures 11.13 and 11.14). Use vector addition to determine the resultant electric field. EAx EBx (1.180 107 N/C)(cos 36.9°) EAy 9.436 106 N/C (1.498 107 N/C)(cos 53.1°) EBy 8.994 106 N/C (1.180 107 N/C)(sin 36.9°) 7.085 106 N/C (1.498 107 N/C)(sin 53.1°) 1.198 107 N/C Add the x components: Enetx EBx EAx (9.436 106 N/C) (8.994 106 N/C) 1.843 107 N/C Add the y components: Enety EBy EAy (7.085 106 N/C) (1.198 107 N/C) 4.895 106 N/C Use the Pythagorean theorem to solve for the magnitude of the electric field: E (1.843 107 N/C)2 (4.895 106 N
/C)2 net 1.91 107 N/C Use the tangent function to determine the direction of the net electric field at point P (Figure 11.15). tan 14.9° The direction of the net field is 180° 14.9° 165° 4.895 106 N/C Enet θ x 1.843 107 N/C Figure 11.15 Paraphrase The net electric field at point P is 1.91 107 N/C [165°]. 552 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 553 In chapter 10, you learned that there are two types of electric charges that interact and are affected by electrostatic forces. In this section, you have learned that these charges are surrounded by electric fields—regions of electric influence around every charge. Electrostatic forces affect charges placed in these fields. Fields explain how two charges can interact, even though there is no contact between them. Since electric fields are vector fields, you can use vector addition to determine a net electric field at a point in the presence of more than one charge in one-dimensional and two-dimensional situations. 11.1 Check and Reflect 11.1 Check and Reflect Knowledge 1. What is the difference between an electric force and an electric field? 2. Why was it necessary to introduce a “field theory”? 3. How is the direction of an electric field defined? 4. Why is an electric field classified as a vector field? 5. If vector arrows can represent an electric field at a point surrounding a charge, identify the two ways that the vector arrows, shown below, represent differences in the electric fields around the two source charges. (a) the magnitude and direction of the electric field at a point 0.300 m to the right of the charge (b) the magnitude and direction of the electric force acting on a positive charge of 2.00 10–8 C placed at the point in (a) 8. A small test sphere with a negative charge of 2.50 C experiences an electrostatic attractive force of magnitude 5.10 10–2 N when it is placed at a point 0.0400 m from another larger charged sphere. Calculate (a) the magnitude and direction of the electric field at this point (b) the magnitude and the sign of charge on the larger charged sphere P E 9. A negative charge of 3.00 mC is 1.20 m to the
right of another negative charge of 2.00 mC. Calculate P E 6. Describe the effect on the electric field at a point (a) if the magnitude of the charge producing the field is halved (b) if the sign of the charge producing the field is changed (c) if the magnitude of the test charge in the field is halved Applications 7. Given a small sphere with a positive charge of 4.50 10–6 C, determine: (a) the net electric field at a point along the same line and midway between the two charges (b) the point along the same line between the two charges where the net electric field will be zero Extension 10. Four similarly charged spheres of 5.00 C are placed at the corners of a square with sides of 1.20 m. Determine the electric field at the point of intersection of the two diagonals of the square. e TEST To check your understanding of forces and fields, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 553 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 554 11.2 Electric Field Lines and Electric Potential In section 11.1, you learned that the electric field from a charge q at a point P can be represented by a vector arrow, as shown in Figure 11.16. The length and direction of the vector arrow represent the magnitude and direction of the electric field (E ) at that point. By measuring the electric force exerted on a test charge at an infinite number of points around a source charge, a vector value of the electric field can be assigned to every point in space around the source charge. This creates a three-dimensional map of the electric field around the source charge (Figure 11.16). Electric Field Lines For many applications, however, a much simpler method is used to represent electric fields. Instead of drawing an infinite number of vector arrows, you can draw lines, called electric field lines, to represent the electric field. Field lines are drawn so that exactly one field line goes through any given point within the field, and the tangent to the field line at the point is in the direction of the electric field vector at that point. You can give the field lines a direction such that the direction of the field line through a given point agrees with the direction of the electric field at that point. Use the following rules when you draw electric field
lines around a point charge: • Electric field lines due to a positive source charge start from the charge and extend radially away from the charge to infinity. • Electric field lines due to a negative source charge come from infin- ity radially into and terminate at the negative source charge. • The density of lines represents the magnitude of the electric field. In other words, the more closely spaced and the greater the number of lines, the stronger is the electric field. Figure 11.16 A threedimensional map of the electric field around a source charge info BIT A lightning rod works because of the concentration of charges on the point of a conductor. This concentration of charge creates an electric field that ionizes air molecules around the point. The ionized region either makes contact with an upward streamer to a cloud, thus preventing the formation of a damaging return lightning stroke, or intercepts a downward leader from the clouds and provides a path for the lightning to the ground to prevent damage to the structure. e SIM Explore the electric fields around a point charge and two charges. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Figure 11.17 shows how to draw electric field lines around one and two negative point charges. Figure 11.17 The field lines around these charges were drawn using the rules given above. 554 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 555 M I N D S O N Drawing Electric Field Lines Rarely is the electric field at a point in space influenced by a single charge. Often, you need to determine the electric field for a complicated arrangement of charges. Electric field lines can be used to display these electric fields. In Figure 11.18, lawn seeds have been sprinkled on the surface of a container of cooking oil. In each case, a different charged object has been put into the oil. • On a sheet of paper, sketch the electric field lines in each situation using the rules for drawing electric field lines given on page 554. • Use concise statements to justify the pattern you drew in each of the sketches. (a) (d) (b) (c) (e) Figure 11.18 (a) one negative charge, (b) two negative charges, (c) one negative and one positive charge, (d) two oppositely charged plates, (e) one negatively charged cylindrical ring Conductors and Electric Field Lines In a conductor, electrons
move freely until they reach a state of static equilibrium. For static equilibrium to exist, all charges must be at rest and thus must experience no net force. Achieving static equilibrium creates interesting distributions of charge that occur only in conducting objects and not in non-conducting objects. Following are five different situations involving charge distribution on conductors and their corresponding electric field lines. Solid Conducting Sphere When a solid metal sphere is charged, either negatively or positively, does the charge distribute evenly throughout the sphere? To achieve static equilibrium, all excess charges move as far apart as possible because of electrostatic forces of repulsion. A charge on the sphere at position A in Figure 11.19(a), for example, would experience a net force of electrostatic repulsion from the other charges. Consequently, all excess charges on a solid conducting sphere are repelled. These excess charges distribute evenly on the surface of the metal conducting sphere. Chapter 11 Electric field theory describes electrical phenomena. 555 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 556 Figure 11.19(b) shows the electric field lines created by the distribution of charge on the surface of a solid conducting sphere. Because electric field lines cannot have a component tangential to this surface, the lines at the outer surface must always be perpendicular to the surface. (a) F F A F F F F (b) Figure 11.19(a) solid sphere Charges on a Figure 11.19(b) for a charged solid sphere Electric field lines Solid, Flat, Conducting Plate How do excess charges, either positive or negative, distribute on a solid, flat, conducting plate like the one in Figure 11.20(a)? On a flat surface, the forces of repulsion are similarly parallel or tangential to the surface. Thus, electrostatic forces of repulsion acting on charges cause the charges to spread and distribute evenly along the outer surface of a charged plate, as shown in Figure 11.20(b). Electric field lines extend perpendicularly toward a negatively charged plate. The electric field lines are uniform and parallel, as shown in Figure 11.20(c). (a) (b) (c) F F F F Figure 11.20 (a) Forces among three charges on the top surface of a flat, conducting plate (b) Uniform distribution of charges on a charged, flat, conducting plate (c) Uniform distribution of charges, shown with electric field lines Irregularly Shaped Solid Conducting Object For
an irregularly shaped solid conductor, the charges are still repelled and accumulate on the outer surface. But do the charges distribute evenly on the outer surface? Figure 11.21(a) is an example of a charged, irregularly shaped object. 556 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 557 On a flatter part of the surface, the forces of repulsion are nearly parallel or tangential to the surface, causing the charges to spread out more, as shown in Figure 11.21(b). At a pointed part of a convex surface, the forces are directed at an angle to the surface, so a smaller component of the forces is parallel or tangential to the surface. With less repulsion along the surface, more charge can accumulate closer together. As a rule, the net electrostatic forces on charges cause the charges to accumulate at the points of an irregularly shaped convex conducting object. Conversely, the charges will spread out on an irregularly shaped concave conducting object. On irregularly shaped conductors, the charge density is greatest where the surface curves most sharply (Figure 11.21(c)). The density of electric field lines is also greatest at these points. info BIT The accumulation of charge on a pointed surface is the explanation for St. Elmo’s fire, which you read about at the beginning of this chapter. St. Elmo’s fire is a plasma (a hot, ionized gas) caused by the powerful electric field from the charge that accumulates on the tips of raised, pointed conductors during thunderstorms. St. Elmo’s fire is known as a form of corona discharge or point discharge. (a) (b) (c) F F y x F F x y Figure 11.21(a) A charged, irregularly shaped convex object Figure 11.21(b) Forces affecting charges on the surface of an irregularly shaped convex object Figure 11.21(c) lines around a charged irregularly shaped convex object Electric field Hollow Conducting Object When a hollow conducting object is charged, either negatively or positively, does the charge distribute evenly throughout the inner and outer surfaces of the object? As you saw in Figures 11.19, 11.20, and 11.21, excess charges move to achieve static equilibrium, and they move as far apart as possible because of electrostatic forces of repulsion. In a hollow conducting object,
all excess charges are still repelled outward, as shown in Figure 11.22(a). However, they distribute evenly only on the outer surface of the conducting object. There is no excess charge on the inner surface of the hollow object, no matter what the shape of the object is. The corresponding electric field lines created by the distribution of charge on the outer surface of a hollow object are shown in Figure 11.22(b). The electric field lines at the outer surface must always be perpendicular to the outer surface. (a) (b) Figure 11.22(a) A charged hollow conducting object Figure 11.22(b) Electric field lines on a hollow conducting object Chapter 11 Electric field theory describes electrical phenomena. 557 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 558 info BIT Coaxial cable wires are used to transmit electric signals such as cable TV to your home. To prevent electric and magnetic interference from outside, a covering of conducting material surrounds the coaxial wires. Any charge applied to the conducting layer accumulates on the outside of the covering. No electric field is created inside a hollow conductor, so there is no influence on the signals transmitted in the wires. e WEB Research the operation of an ink-jet printer. What is the function of charged plates in these printers? Begin your search at www.pearsoned.ca/school/ physicssource. Most surprisingly, the electric field is zero everywhere inside the conductor, so there are no electric field lines anywhere inside a hollow conductor. As previously described, this effect can be explained using the superposition principle. Fields set up by many sources superpose, forming a single net field. The vector specifying the magnitude of the net field at any point is simply the vector sum of the fields of each individual source. Anywhere within the interior of a hollow conducting object, the vector sum of all the individual electric fields is zero. For this reason, the person inside the Faraday cage, shown in the photograph on page 508, is not affected by the tremendous charges on the outside surface of the cage. Parallel Plates If two parallel metal plates, such as those in Figure 11.23(a), are oppositely charged, how are the charges distributed? Electrostatic forces of repulsion of like charges, within each plate, cause the charges to distribute evenly within each plate, and electrostatic forces of attraction of opposite charges on the two plates cause the charges to accumulate on the inner surfaces. Thus
, the charges spread and distribute evenly on the inner surfaces of the charged plates. (a) (b) Figure 11.23(a) The distribution of net charge on oppositely charged parallel plates Figure 11.23(b) Electric field lines between two oppositely charged parallel plates The magnitude of the resulting electric field can be shown to be the vector sum of each individual field, so it can be shown that the electric field anywhere between the plates is uniform. Thus, between two oppositely charged and parallel plates, electric field lines exist only between the charged plates. These lines extend perpendicularly from the plates, starting at the positively charged plate and terminating at the negatively charged plate. The electric field lines are uniform in both direction and density between the two oppositely charged plates, except near the edges of the plates. Such a system is called a parallel-plate capacitor. This type of capacitor is found in many different types of electrical equipment, including printers and televisions (where it is part of the “instant on” feature). It is also used in particle accelerators, such as cathode-ray tubes and mass spectrometers. You will learn about mass spectrometers in Unit VIII. 558 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 559 THEN, NOW, AND FUTURE Defibrillators Save Lives During a heart attack, the upper and lower parts of the heart can begin contracting at different rates. Often these contractions are extremely rapid. This fluttery unsynchronized beating, called fibrillation, pumps little or no blood and can damage the heart. A defibrillator uses a jolt of electricity to momentarily stop the heart so that it can return to a normal beat (Figure 11.24). Figure 11.24 A defibrillator stops the fibrillation of the heart muscle by applying an electric shock. A defibrillator consists of two parallel charged plates (see Figure 11.23(b)), called a parallel-plate capacitor, connected to a power supply and discharging pads. A typical defibrillator stores about 0.4 C on the plates, creating a potential difference of approximately 2 kV between the plates. When discharged through conductive pads placed on the patient’s chest, the capacitor delivers about 0.4 kJ of electrical energy in 0.002 s. Roughly 200 J of this energy passes through the patient’s chest.
A defibrillator uses a highvoltage capacitor to help save lives. Such capacitors have many other applications in other electrical and electronic devices, such as the highvoltage power supplies for cathoderay tubes in older televisions and computer monitors. The charge stored in such capacitors can be dangerous. Products con- taining such high-voltage capacitors are designed to protect the users from any dangerous voltages. However, service technicians must be careful when working on these devices. Since the capacitors store charge, they can deliver a nasty shock even after the device is unplugged. Questions 1. How does the magnitude of the power delivered by the plates compare with the actual power delivered to the chest by the jolt? 2. Identify a feature of televisions that demonstrates an important application of parallel-plate capacitors. 3. If a defibrillator can store 0.392 C of charge in 30 s, how many electrons are stored in this time period? M I N D S O N Faraday’s Ice Pail In the early 1800s, Michael Faraday performed an experiment to investigate the electric fields inside a hollow metal container. He used ice pails, so this experiment is often called “Faraday’s ice pail experiment.” charged rod This activity is called a conceptual experiment because you will not perform the experiment. Instead, you will predict and justify the results of an experimental procedure that duplicates Faraday’s investigation. The purpose of the experiment is to determine what type of electric field exists on the inside and the outside of a hollow metal container. A positively charged rod is placed into position inside the metal container, near the centre, as shown in Figure 11.25. The rod is then moved to a position inside the metal container, near one of the inner surfaces. • Which of the electroscopes would show a deflection when the rod is near the centre of the metal container? • Clearly explain your reasoning and the physical principles you used in determining your answers to these questions. insulated plate Figure 11.25 An ice pail is a metal container. It is placed on an insulated surface, and electroscopes are attached to the inside and outside surfaces of the metal container. Chapter 11 Electric field theory describes electrical phenomena. 559 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 560 Figure 11.26 The charged dome of a Van de Graaff generator exposes a person to very
large voltages. Electric Potential Energy and Electric Potential A Van de Graaff generator can generate up to 250 kV. Touching the dome not only produces the spectacular results shown in Figure 11.26, it can also cause a mild, harmless shock. On the other hand, touching the terminals of a wall socket, which has a voltage of 120 V, can be fatal. An understanding of this dramatic difference between the magnitude of the voltage and its corresponding effect requires a study of the concepts of electric potential energy and electric potential. These concepts are important in the study of electric fields. Even though the terms seem similar, they are very different. To explain the difference, you will study these concepts in two types of electric fields: non-uniform electric fields around point charges, and uniform electric fields between parallel charged plates. Electric Potential Energy In previous grades, you learned about the relationship between work and potential energy. Work is done when a force moves an object in the direction of the force such that: d W F where W is work, and F displacement of the object. and d are the magnitudes of the force and the In a gravitational system like the one shown in Figure 11.27(a), lifting a mass a vertical distance against Earth’s gravitational field requires work to stretch an imaginary “gravitational spring” connecting the mass and Earth. Further, because the force required to do the work is a conservative force, the work done against the gravitational field increases the gravitational potential energy of the system by an amount equal to the work done. Therefore: gravitational potential energy gain work done (a) Fapp d mass Fg Ep W (b) Fapp d Fe Figure 11.27(a) Work is required to lift a mass to a certain position above Earth’s surface. Figure 11.27(b) Work is required to move a small positive charge away from a larger negative charge. 560 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 561 Similarly, in an electrostatic system like the one shown in Figure 11.27(b), moving a small charge through a certain distance in a non-uniform electric field produced by another point charge requires work to either compress or stretch an imaginary “electrostatic spring” connecting the two charges. Since the force required to do this work is also a conservative force, the work done in the electric field must increase the electric potential energy of the system. Electric
potential energy is the energy stored in the system of two charges a certain distance apart (Figure 11.28). Electric potential energy change equals work done to move a small charge: Ep W q1 r P q2 Figure 11.28 Electric potential energy is the energy stored in the system of two charges a certain distance apart. Example 11.5 Moving a small charge from one position in an electric field to another position requires 3.2 10–19 J of work. How much electric potential energy will be gained by the charge? Analysis and Solution The work done against the electrostatic forces is W. The electric potential energy gain is Ep. In a conservative system, Ep W So, Ep W 3.2 1019 J The electric potential energy gain of the charge is 3.2 10–19 J. Practice Problems 1. A small charge gains 1.60 10–19 J of electric potential energy when it is moved to a point in an electric field. Determine the work done on the charge. 2. A charge moves from one position in an electric field, where it had an electric potential energy of 6.40 10–19 J, to another position where it has an electric potential energy of 8.00 10–19 J. Determine the work necessary to move the charge. Answers 1. 1.60 1019 J 2. 1.60 1019 J Choosing a Reference Point In Chapter 7, you learned that commonly used reference points for zero gravitational potential energy are Earth’s surface or infinity. Choosing a zero reference point is necessary so you can analyze the relationship between work and gravitational potential energy. Chapter 11 Electric field theory describes electrical phenomena. 561 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 562 Consider a zero reference point at Earth’s surface. An object at rest on Earth’s surface would have zero gravitational potential energy relative to Earth’s surface. If the object is lifted upward, opposite to the direction of the gravitational force it experiences, then work is being done on the object. The object thus gains gravitational potential energy. If the object falls back to the surface in the same direction as the gravitational force, then the object loses gravitational potential energy. As with gravitational potential energy, the value of electric potential energy at a certain position is meaningless unless it is compared to a reference point where the electric potential energy is zero. The choice of a zero reference point for electric potential energy is arbitrary. For example, suppose an electric
field is being produced by a large negative charge. A small positive charge would be attracted and come to rest on the surface of the larger negative charge, where it would have zero electric potential energy. This position could be defined as a zero electric potential energy reference point (Figure 11.29(a)). Then, the test charge has positive electric potential energy at all other locations. Alternatively, the small positive test charge may be moved to a position so far away from the larger negative charge that there is no electrostatic attraction between them. This position would be an infinite distance away. This point, at infinity, is often chosen as the zero electric potential energy reference point. Then, the test charge has negative electric potential energy at all other locations. This text uses infinity as the zero electric potential energy reference point for all calculations (Figure 11.29(b)). at surface at infinity q q Ep 0 (a) Ep 0 (b) Figure 11.29 Two commonly used reference points for electric potential energy: (a) test charge defined as having zero electric potential energy at the surface of the source charge (b) test charge defined as having zero electric potential energy at infinity 562 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 563 Work and Electric Potential Energy Whenever work is done on a charge to move it against the electric force caused by an electric field, the charge gains electric potential energy. The following examples illustrate the relationship between work and electric potential energy. Electric Potential Energy Between Parallel Charged Plates Except at the edges, the electric field between two oppositely charged plates is uniform in magnitude and direction. Suppose a small positive charge in the field between the plates moves from the negative plate to the positive plate with a constant velocity. This motion requires an external force to overcome the electrostatic forces the charged plates exert on the positive charge. The work done on the charge increases the system’s electric potential energy: Ep W F d Example 11.6 When a small positive charge moves from a negative plate to a positive plate, 2.3 × 10–19 J of work is done. How much electric potential energy will the charge gain? Analysis and Solution In a conservative system, EP EP W 2.3 1019 J W. Paraphrase The electric potential energy gain of the charge is 2.3 1019 J. Practice Problem 1. A charge gained 4.00 105 J of electric potential energy when it was moved between two oppositely charged plates.
How much work was done on the charge? Answer 1. 4.00 105 J Electric Potential Suppose two positive charges are pushed toward a positive plate. In this case, twice as much work is done, and twice as much electric potential energy is stored in the system. However, just as much electric potential energy is still stored per charge. Storing 20 J of energy in two charges is the same as storing 10 J of energy in each charge. At times, it is necessary to determine the total electric potential energy at a certain location in an electric field. At other times, it is convenient to consider just the electric potential energy per unit charge at a location. The electric potential energy stored per unit charge at a given point is the amount of work required to move a unit charge to that Chapter 11 Electric field theory describes electrical phenomena. 563 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 564 electric potential: the electric potential energy stored per unit charge at a given point in an electric field point from a zero reference point (infinity). This quantity has a special name: electric potential. To determine the electric potential at a location, use this equation: electric potential electric potential energy charge Ep V q where V is in volts, Ep is in joules, and q is in coulombs. Since electric potential energy is measured in joules and charge is measured in coulombs, o e l u j 1 volt b m o l u co 1 1 Thus, if the electric potential at a certain location is 10 V, then a charge of 1 C will possess 10 J of electric potential energy, a charge of 2 C will possess 20 J of electric potential energy, and so on. Even if the total electric potential energy (Ep) at a location changes, depending on the amount of charge placed in the electric field, the electric potential (V ) at that location remains the same. A balloon can be used as an example to help explain the difference between the concepts of electric potential energy and electric potential. Suppose you rub a balloon with fur. The balloon acquires an electric potential of a few thousand volts. In other words, the electric energy stored per coulomb of charge on the balloon is a few thousand volts. Written as an equation, Ep V q Now suppose the balloon were to gain a large charge of 1 C during the rubbing process. In order for the electric potential to stay the same, a few thousand joules of work would be needed to
produce the electrical energy that would allow the balloon to maintain that electric potential. However, the amount of charge a balloon acquires during rubbing is usually only in the order of a few microcoulombs. So, acquiring this potential requires a small amount of work to produce the energy needed. Even though the electric potential is high, the electric potential energy is low because of the extremely small charge. Concept Check Suppose the magnitude of a charge placed in an electric field were doubled. How much would the electric potential energy and the electric potential change? info BIT The SI unit of electric potential is the volt, named in honour of the Italian physicist Count Alessandro Volta (1745–1827), who developed the first electric battery in the early 1800s. 564 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 565 Electric Potential Difference When a charge moves from one location to another in an electric field, it experiences a change in electric potential. This change in electric potential is called the electric potential difference, V, between the two points and V Vfinal Vinitial Ep since V q Ep V q where Ep is the amount of work required to move the charge from one location to the other. The potential difference depends only on the two locations. It does not depend on the charge or the path taken by the charge as it moves from one location to another. Electric potential difference is commonly referred to as just potential difference or voltage. An electron volt (eV) is the quantity of energy an electron gains or loses when passing through a potential difference of exactly 1 V. An electron volt is vastly less than a joule: 1 eV 1.60 1019 J Although not an SI unit, the electron volt is sometimes convenient for expressing tiny quantities of energy, especially in situations involving a single charged particle such as an electron or a proton. The energy difference in Example 11.6 could be given as (2.3 1019 J) 1.4 eV V e 1 019 J 1 1.60 electric potential difference: change in electric potential experienced by a charge moving between two points in an electric field electron volt: the change in energy of an electron when it moves through a potential difference of 1 V PHYSICS INSIGHT The notation VAB is widely used instead of V to represent the potential difference at point A relative to point B. When the points in question are clear from the context, the subscripts are generally omitted. For example
, the equation for Ohm’s law is usually written as V IR, where it is understood that V represents the potential difference between the ends of the resistance R. Example 11.7 Moving a small charge of 1.6 × 10–19 C between two parallel plates increases its electric potential energy by 3.2 × 10–16 J. Determine the electric potential difference between the two parallel plates. Analysis and Solution To determine the electric potential difference between the plates, use the equation Ep.0 103 V The electric potential difference between the plates is 2.0 103 V. Practice Problems 1. In moving a charge of 5.0 C from one terminal to the other, a battery raises the electric potential energy of the charge by 60 J. Determine the potential difference between the battery terminals. 2. A charge of 2.00 10–2 C moves from one charged plate to an oppositely charged plate. The potential difference between the plates is 500 V. How much electric potential energy will the charge gain? Answers 1. 12 V 2. 10.0 J Chapter 11 Electric field theory describes electrical phenomena. 565 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 566 Example 11.8 A small charge of 3.2 1019 C is moved between two parallel plates from a position with an electric potential of 2.0 103 V to another position with an electric potential of 4.0 103 V (Figure 11.30). Practice Problems 1. A sphere with a charge of magnitude 2.00 C is moved between two positions between oppositely charged plates. It gains 160 J of electric potential energy. What is the potential difference between the two positions? 2. An electron moves between two positions with a potential difference of 4.00 104 V. Determine the electric potential energy gained by the electron, in joules (J) and electron volts (eV). Answers 1. 80.0 V 2. 6.40 1015 J or 4.00 104 eV A 2.0 103 V B 4.0 103 V Figure 11.30 3.2 1019 C battery Determine: (a) the potential difference between the two positions (b) the electric potential energy gained by moving the charge, in joules (J) and electron volts (eV) Given Vinitial Vfinal 2.0 103 V 4.0 103 V q 3.2 10–19 C Required (a) potential difference between points B and A (
V ) (b) electric potential energy gained by the charge (Ep) Analysis and Solution (a) V Vfinal Vinitial (4.0 103 V) (2.0 103 V) 2.0 103 V (b) To calculate the electric potential energy, use the equation Ep. V q Vq (2.0 103 V)(3.2 1019 C) 6.4 10–16 J Ep Ep Since 1 eV 1.60 1019 J, V e 1 (6.4 1016 J) 019 J 1 4.0 103 eV 4.0 keV 1.60 Paraphrase (a) The potential difference between the two positions is 2.0 103 V. (b) The energy gained by moving the charge between the two positions is 6.4 10–16 J or 4.0 103 eV. 566 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 567 The Electric Field Between Charged Plates Earlier in this section, you determined the electric field strength surrounding a point charge using the following equations: k q or E E 2 r F e q You also learned that the electric field around a point charge is a nonuniform electric field. Its magnitude depends on the distance from the charge. Later, you learned that a special type of electric field exists between two charged parallel plates. The magnitude of the electric field between the plates is uniform anywhere between the plates and it can be determined using the general equation for an electric field, F. You cannot use the equation E E e q k q because it is used only 2 r e WEB One of the technological applications of parallelplate capacitors is in disposable cameras. Research the role of capacitors in these cameras. Begin your search at www.pearsoned.ca/ school/physicssource. for point charges. Now, after studying electric potential difference, you can see how another equation for determining the electric field strength between plates arises from an important relationship between the uniform electric field and the electric potential difference between two charged parallel plates (Figure 11.31). If a small positively charged particle (q) is moved through the uni- form electric field (E q. This force is ), a force is required, where F the force exerted on the particle due to the presence of the electric field. If this force moves the charged particle a distance (d) between the plates, then the work done is: E
Figure 11.31 Electrically charged parallel plates W F or W E d qd Since this system is conservative, the work done is stored in the charge as electric potential energy: W Ep E qd The electric potential difference between the plates is: Ep V q q d E q d E To calculate the magnitude of the uniform electric field between charged plates, use the equation E V d where V is the electric potential difference between two charged plates in volts; d is the distance in metres between the plates; and E is the magnitude of the electric field in volts per metre. Chapter 11 Electric field theory describes electrical phenomena. 567 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 568 Note that 1 V/m equals 1 N/C because /C J 1 V//C Example 11.9 Practice Problems 1. Two charged parallel plates, separated by 5.0 10–4 m, have an electric field of 2.2 104 V/m between them. What is the potential difference between the plates? 2. Spark plugs in a car have electrodes whose faces can be considered to be parallel plates. These plates are separated by a gap of 5.00 10–3 m. If the electric field between the electrodes is 3.00 106 V/m, calculate the potential difference between the electrode faces. Answers 1. 11 V 2. 1.50 104 V A cathode-ray-tube (CRT) computer monitor accelerates electrons between charged parallel plates (Figure 11.32). These electrons are then directed toward a screen to create an image. If the plates are 1.2 10–2 m apart and have a potential difference of 2.5 104 V between them, determine the magnitude of the electric field between the plates. 1.2 102 m V 2.5 104 V accelerating plates Given V 2.5 104 V d 1.2 10–2 m screen Figure 11.32 Required ) magnitude of the electric field between the plates (E Analysis and Solution To calculate the magnitude of the electric field between the plates, use the equation V E d 4 V 0 1.5 2 2 1 1 m.2 2.1 106 V/m 0 Paraphrase The magnitude of the electric field between the plates is 2.1 106 V/m. 568 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 569 11.2 Check and Reflect 11.2 Check
and Reflect Knowledge 1. Describe the difference between an electric field vector and an electric field line. 2. Sketch electric field lines around the following charges: (a) a positive charge (b) a negative charge (c) two positive charges (d) two negative charges (e) a positive charge and a negative charge 3. Describe the difference between electric potential and electric potential energy. Applications 4. At a point in Earth’s atmosphere, the electric field is 150 N/C downward and the gravitational field is 9.80 N/kg downward. (a) Determine the electric force on a proton (p) placed at this point. (b) Determine the gravitational force on the proton at this point. The proton has a mass of 1.67 10–27 kg. 9. Determine the magnitude and direction of the net electric field at point P shown in the diagram below. 50 μC 10 μC P 0.30 m 0.15 m 10. A uniform electric field exists between two oppositely charged parallel plates connected to a 12.0-V battery. The plates are separated by 6.00 104 m. (a) Determine the magnitude of the electric field between the plates. (b) If a charge of 3.22 106 C moves from one plate to another, calculate the change in electric potential energy of the charge. Extensions 11. A metal car is charged by contact with a charged object. Compare the charge distribution on the outside and the inside of the metal car body. Why is this property useful to the occupants of the car if the car is struck by lightning? 5. A metal box is charged by touching it with 12. Explain why only one of the electroscopes a negatively charged object. (a) Compare the distribution of charge at the corners of the box with the faces of the box. (b) Draw the electric field lines inside and surrounding the box. 6. What is the electric field intensity 0.300 m away from a small sphere that has a charge of 1.60 10–8 C? connected to the hollow conductive sphere in the illustration below indicates the presence of a charge. 7. Calculate the electric field intensity 13. Two points at different positions in midway between two negative charges of 3.2 C and 6.4 C separated by 0.40 m. 8. A 2.00-C charge jumps across a spark gap in a spark plug across which the potential difference is 1.00 103 V.
How much energy is gained by the charge? an electric field have the same electric potential. Would any work be required to move a test charge from one point to another? Explain your answer. e TEST To check your understanding of electric field lines, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 569 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 570 info BIT Living cells “pump” positive sodium ions (Na) from inside a cell to the outside through a membrane that is 0.10 m thick. The electric potential is 0.70 V higher outside the cell than inside it. To move the sodium ions, work must be done. It is estimated that 20% of the energy consumed by the body in a resting state is used to operate these “pumps.” 11.3 Electrical Interactions and the Law of Conservation of Energy A charge in an electric field experiences an electrostatic force. If the charge is free to move, it will accelerate in the direction of the electrostatic force, as described by Newton’s second law. The acceleration of the charge in the non-uniform electric field around a point charge is different from the acceleration motion of a charge in a uniform electric field between charged plates. Figure 11.33 shows a charge in the non-uniform field of a point charge. The electrostatic force on a charge placed in the field varies inversely as the square of the distance between the charges. A varying force causes non-uniform acceleration. Describing the motion of the charge in this type of situation requires applying calculus to Newton’s laws of motion, which is beyond the scope of this text. However, to determine the particle’s speed at a given point, you can use the law of conservation of energy. source charge F a E Figure 11.33 The electrostatic force on a point charge in a non-uniform electric field causes non-uniform acceleration of the charge. If the forces acting on an object are conservative forces, then the work done on a system changes the potential energy of the system. Electric potential energy, like gravitational potential energy, can be converted to kinetic energy. A charged particle placed in an electric field will accelerate from a region of high potential energy to a region of low potential energy. According to the law of conservation of energy, the moving charge gains kinetic energy at
the expense of potential energy. If you assume that no energy is lost to friction and the forces are conservative, the kinetic energy gained equals the potential energy lost, so the sums of the two energies are always equal: Epi Eki Epf Ekf 570 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:38 AM Page 571 Example 11.10 A pith ball of mass 2.4 104 kg with a positive charge of 1.2 108 C is initially at rest at location A in the electric field of a larger charge (Figure 11.34). At this location, the charged pith ball has 3.0 107 J of electric potential energy. When released, the ball accelerates toward the larger charge. At position B, the ball has 1.5 108 J of electric potential energy. Find the speed of the pith ball when it reaches position B. 1.2 108 C 2.4 104 kg B 1.5 108 J A 3.0 107 J Figure 11.34 Given m 2.4 104 kg 3.0 107 J Epi q 1.2 108 C 1.5 108 J Epf Required speed of the ball at position B (v) Analysis and Solution The pith ball is at rest at A, so its initial kinetic energy is zero. Its electric potential energy at B is lower than at A. Since this system is conservative, the loss of electric potential energy when the ball moves from A to B is equal to a gain in kinetic energy, according to the law of conservation of energy: Epf Ekf Eki Epi Substitute the given values and solve for Ekf (3.0 107 J) 0 (1.5 108 J) Ekf Ekf Since the kinetic energy of an object is Ek 2.85 107 J. 1 mv2, 2 Practice Problems 1. A negative charge of 3.00 10–9 C is at rest at a position in the electric field of a larger positive charge and has 3.20 1012 J of electric potential energy at this position. When released, the negative charge accelerates toward the positive charge. Determine the kinetic energy of the negative charge just before it strikes the larger positive charge. 2. A small sphere with a charge of 2.00 C and a mass of 1.70 103 kg accelerates from rest toward a larger positive charge. If the speed of the sphere just before it strikes the positive charge is 5.20 104 m/s
, how much electric potential energy did the negative charge lose? Answers 1. 3.20 1012 J 2. 2.30 106 J Ek 2 v 2 m v 2Ek m 2(2.85 107 J) 2.4 104 kg 4.9 102 m/s Paraphrase The speed of the pith ball at position B is 4.9 102 m/s. Chapter 11 Electric field theory describes electrical phenomena. 571 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 572 It is easier to describe the motion of a charge in a uniform electric field between two parallel plates, as shown in Figure 11.35. In this case, the acceleration is constant because of the constant force, so either the work–energy theorem or the laws of dynamics can be used. (Because the electric field is constant (uniform), the force acting on a charge q is also constant because F e qE.) F E E Figure 11.35 In a uniform electric field between two parallel plates, the acceleration of a charge is constant. Concept Check Electrostatic forces and gravitational forces are similar, so the motion of objects due to these forces should be similar. Consider a charge in an electric field between two parallel plates. Sketch the direction of the motion of the charge when its initial motion is: • perpendicular to the plates (the electrostatic force is similar to the gravitational force on falling masses) • parallel to the plates (the electrostatic force is similar to the gravitational force that causes the parabolic projectile motion of a mass close to the surface of a large planet or moon) Example 11.11 Two vertical parallel plates are connected to a DC power supply, as shown in Figure 11.36. The electric potential between the plates is 2.0 103 V. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is placed at the positive plate and released. It accelerates toward the negative plate. Determine the speed of the sphere at the instant before it strikes the negative plate. Ignore any gravitational effects. 2.6 1012 C 3.0 1015 kg 2.0 103 V Figure 11.36 572 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 573 Given q 2.6 1012 C V 2.0 103 V m 3.0 1015 kg Required speed of the sphere at the negative plate (v)
Analysis and Solution This system is conservative. You can use kinetic energy of the charge to find its speed., is 0 J. 0 J. The initial electric potential energy of the sphere at the positive plate Vq. Since the sphere was at rest, its initial kinetic energy, is Epi Eki The final electric potential energy of the sphere at the negative plate is Epf According to the law of conservation of energy, Ekf Epi Vq 0 J 0 J Ekf (2.0 103 V)(2.6 1012 C) 0 J 0 J Ekf Epf Eki 5.2 109 J Ekf Since Ek 1 mv 2, 2 v 2Ek m 2(5.2 109 J) 3.0 1015 kg 1.9 103 m/s Paraphrase The speed of the sphere at the negative plate is 1.9 103 m/s. Practice Problems 1. An alpha particle with a charge of 3.20 1019 C and a mass of 6.65 1027 kg is placed between two oppositely charged parallel plates with an electric potential difference of 4.00 104 V between them. The alpha particle is injected at the positive plate with an initial speed of zero, and it accelerates toward the negative plate. Determine the final speed of the alpha particle just before it strikes the negative plate. 2. If a charge of 6.00 106 C gains 3.20 104 J of kinetic energy as it accelerates between two oppositely charged plates, what is the potential difference between the two parallel plates? Answers 1. 1.96 106 m/s 2. 53.3 V Chapter 11 Electric field theory describes electrical phenomena. 573 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 574 Example 11.12 An electron enters the electric field between two charged parallel plates, as shown in Figure 11.37. electric field Figure 11.37 (a) Copy Figure 11.37 into your notebook and sketch the motion of the electron between the plates. (b) If the electron experiences a downward acceleration of 2.00 1017 m/s2 due to the electric field between the plates, determine the time taken for the electron to travel 0.0100 m to the positive plate. Given a 2.00 1017 m/s2 [down] d 0.0100 m Required (a) sketch of the electron’s motion (b) time (t) Analysis and Solution (a) The electron
’s acceleration is downward, so the motion of the electron will follow a parabolic path to the positive plate (Figure 11.38), similar to the projectile motion of an object travelling horizontally to the surface of Earth and experiencing downward acceleration due to gravity. electric field Figure 11.38 Practice Problems 1. Two horizontal parallel plates, 1.2 102 m apart, are connected to a DC power supply, as shown in the figure below. The electric field between the plates is 1.7 105 V/m. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is injected into the region between the plates, with an initial speed of 3.3 103 m/s, as shown. It accelerates toward the negative plate. Copy the diagram into your notebook, sketch the motion of the positive charge through the region between the plates, and determine the distance the positive charge moves toward the negative plate after 6.0 106 s have elapsed. Gravitational effects may be ignored in this case. electric field 1.7 105 V/m 3.3 103 m/s 2. An electron, travelling at 2.3 103 m/s, enters perpendicular to the electric field between two horizontal charged parallel plates. If the electric field strength is 1.5 102 V/m, calculate the time taken for the electron to deflect a distance of 1.0 10–2 m toward the positive plate. Ignore gravitational effects. Answers 1. 2.7 103 m 2. 2.7 108 s 574 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:40 AM Page 575 (b) Use the equation d vi t 1 a(t)2 to determine the 2 time it takes the electron to fall to the positive plate. Since vi d 1 a(t)2 2 t 2d a 2(0.0100 m) m 2.00 1017 s2 0, 3.16 1010 s Paraphrase (a) The path of the electron between the parallel plates is parabolic. (b) The time taken for the electron to fall to the positive plate is 3.16 1010 s. 11.3 Check and Reflect 11.3 Check and Reflect Knowledge Applications 1. In what direction will a positively charged particle accelerate in an electric field? 2. Electric potential energy exists only where a charge is present at a point in an electric field. Must a charge also be present at that point
for there to be electric potential? Why or why not? 4. Calculate the speed of an electron and a proton after each has accelerated from rest through an electric potential of 220 V. 5. Electrons in a TV picture tube are accelerated by a potential difference of 25 kV. Find the maximum speed the electrons would reach if relativistic effects are ignored. 3. Two positively charged objects are an 6. A charge gains 1.92 1014 J of electric equal distance from a negatively charged object, as shown in the diagram below. Charge B is greater than charge A. Compare the electric potential and electric potential energy of the positively charged objects. B A potential energy when it moves through a potential difference of 3.20 104 V. What is the magnitude of the charge? 7. How much work must be done to increase the electric potential of a charge of 2.00 106 C by 120 V? 8. A deuterium ion (H1), a heavy isotope of hydrogen, has a charge of 1.60 1019 C and a mass of 3.34 1027 kg. It is placed between two oppositely charged plates with a voltage of 2.00 104 V. Find the final maximum speed of the ion if it is initially placed at rest (a) at the positive plate (b) midway between the two plates Chapter 11 Electric field theory describes electrical phenomena. 575 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 576 9. A small charge of +3.0 108 C with a mass of 3.0 105 kg is slowly pulled through a potential difference of 6.0 102 V. It is then released and allowed to accelerate toward its starting position. Calculate (a) the initial work done to move the charge (b) the maximum kinetic energy of the returning charge (c) the final speed of the returning charge 10. An electron, travelling horizontally at a speed of 5.45 106 m/s, enters a parallelplate capacitor with an electric field of 125 N/C between the plates, as shown in the figure below. Extensions 11. Determine whether an electron or a proton would take less time to reach one of a pair of oppositely charged parallel plates when starting from midway between the plates. Explain your reasoning. 12. How can the electric potential at a point in an electric field be high when the electric potential energy is low? 13. In question 10, explain why the resulting motion of an electron,
initially travelling perpendicular to the uniform electric field between the two charged parallel plates, will be parabolic and not circular. e TEST 5.45 106 m/s To check your understanding of electrical interactions and the law of conservation of energy, follow the eTest links at www.pearsoned.ca/ school/physicssource. (a) Copy the diagram into your notebook and sketch (i) the electric field lines between the plates (ii) the motion of the electron through the capacitor (b) Determine the force due to the electric field on the electron. (c) Ignoring gravitational effects, calculate the acceleration of the electron. (d) If the electron falls a vertical distance of 6.20 103 m toward the positive plate, how far will the electron travel horizontally between the plates? 576 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:42 AM Page 577 CHAPTER 11 SUMMARY Key Terms and Concepts field test charge source charge electric field line Key Equations F e E q Ep V q k E r q 2 Ep V q Conceptual Overview electric potential energy electric potential (voltage) electron volt electric potential difference Ep W Ep W F d V Vfinal Vinitial E V d Epi Eki Epf Ekf The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. E Fe q E kq r2 electric potential energy calculate define magnitude direction calculate E between more than two charges one-dimensional situations vector electric potential Electric Fields electric potential difference field lines electric field between plates relationship between electric field and distance drawing electric fields around charged objects define a reference point calculate electric potential energy define calculate define calculate Chapter 11 Electric field theory describes electrical phenomena. 577 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 578 CHAPTER 11 REVIEW Knowledge 1. (11.1) Identify the three theories that attempt to explain “action at a distance.” 13. (11.3) Describe the key differences between the electric field surrounding a point charge and the electric field between charged parallel plates? 2. (11.1) How can it be demonstrated that the space around a charged object is different from the space around an uncharged object? 3. (11.1) How does a vector arrow represent both 14. (11.3) Assuming forces in a system are conservative
, explain how (a) work done in the system is related to potential energy of the system the magnitude and direction of a vector quantity? (b) the kinetic and potential energy of the 4. (11.2) What is the difference between an electric field vector and an electric field line? 5. (11.2) Two hollow metal objects, with shapes shown below, are charged with a negatively charged object. In your notebook, sketch the distribution of charge on both objects and the electric field lines surrounding both objects. cross-section of hollow sphere cross-section of hollow egg-shaped object system are related Applications 15. Compare the electric potential energy of a positive test charge at points A and B near a charged sphere, as shown below. A B 16. A large metal coffee can briefly contacts a charged object. Compare the results when uncharged electroscopes are touched to the inside and outside surfaces of the can. 6. (11.2) How do electric field lines represent the magnitude of an electric field? 17. A point charge has a charge of 2.30 C. Calculate (a) the electric field at a position 2.00 m from 7. (11.2) Where do electric field lines originate for (a) a negative point charge? (b) a positive point charge? 8. (11.2) Identify two equations that can be used to calculate the magnitude of an electric field around a point charge. the charge (b) the electric force on a charge of 2.00 C placed at this point 18. A charge of 5.00 C is separated from another charge of 2.00 C by a distance of 1.20 m. Calculate (a) the net electric field midway between the 9. (11.2) When do electric charges achieve static two charges equilibrium in a charged object? 10. (11.2) Why do electric charges accumulate at a point in an irregularly shaped object? 11. (11.2) State a convenient zero reference point for electric potential energy (a) around a point charge (b) between two oppositely charged parallel plates 12. (11.2) What equation would you use to calculate the electric potential energy at a certain position around a point charge? (b) the position where the net electric field is zero 19. Find the net electric field intensity at point C in the diagram below. C 0.040 m 2.0 μC A 0.060 m 2.0 μC B 578 Unit
VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 579 20. A force of 15.0 N is required to move a charge of 2.0 C through a distance of 0.20 m in a uniform electric field. (a) How much work is done on the charge? (b) How much electric potential energy does the charge gain in joules? 21. How much electric potential energy would an object with a charge of 2.50 C have when it is 1.20 m from a point charge of 3.00 C? (Hint: Consider how much electric potential energy the negatively charged object would have when touching the point charge.) 22. Two parallel plates are separated by a distance of 3.75 cm. Two points, A and B, lie along a perpendicular line between the parallel plates and are 1.10 cm apart. They have a difference in electric potential of 6.00 V. (a) Calculate the magnitude of the electric field between the plates. (b) Determine the electric potential between the parallel plates. 23. How much work is required to move a charge perpendicular to the electric field between two oppositely charged parallel plates? 24. A cell membrane is 1.0 107 m thick and has an electric potential difference between its surfaces of 0.070 V. What is the electric field within the membrane? 25. A lithium nucleus (Li3) that has a charge of 4.80 1019 C is accelerated by a voltage of 6.00 105 V between two oppositely charged plates. Calculate the energy, in joules (J) and electron volts (eV), gained by the nucleus. 26. How much electric potential energy, in joules (J) and electron volts (eV), does an alpha particle gain when it moves between two oppositely charged parallel plates with a voltage of 20 000 V? 27. Consider a sphere with a known charge in the electric field around a larger unknown charge. What would happen to the electric field at a point if Extensions 28. Explain why electric field lines can never cross. 29. A bird is inside a metal birdcage that is struck by lightning. Is the bird likely to be harmed? Explain. 30. Explain why charge redistributes evenly on the outside surface of a spherical charged object and accumulates at a point on an irregularly shaped charged object. 31. Why can there never be excess charges inside a charged conductive sphere
? 32. Describe a simple experiment to demonstrate that there are no excess charges on the inside of a hollow charged sphere. 33. Identify a technology that uses the principle that electric charges accumulate at the point of an irregularly shaped object. Describe how the technology applies this principle. Consolidate Your Understanding Create your own summary of electric field theory by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 577 and the Learning Outcomes on page 542. 1. Create a flowchart to describe the differences between electric fields, electric potential energy, and electric potential, using non-uniform and uniform electric fields. 2. Write a paragraph comparing the electric fields around various objects and surfaces. Include diagrams in your comparisons. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 543. How would you answer each question now? (a) the magnitude of the test charge were doubled? (b) the magnitude of the charge producing the e TEST field were doubled? (c) the sign of the charge producing the field were changed? To check your understanding of concepts presented in Chapter 11, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 579 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 580 Properties of electric and magnetic fields apply in nature and technology. Figure 12.1 Aurora borealis or northern lights The spectacular aurora borealis paints the night sky with shimmering colours in northern latitudes (Figure 12.1). Frequently seen above 60° north, its scientific name translates from Latin into “dawn of the north.” In southern latitudes, where it is seen mainly above 60° south, it is called the aurora australis — “dawn of the south.” Many ancient civilizations created stories to explain these dancing lights in the sky. Some Inuit peoples of northern Canada believed that the sky was a hard dome that arched over Earth. Spirits could pass through a hole in the dome to the heavens, where they would light torches to guide new arrivals. Other Aboriginal traditions spoke of the creator of Earth travelling to the north when he finished his task of creation. There he remained, building large fires to remind his people that he still thought of them.
The northern lights were reflections of these fires. What are the auroras and what causes them? Why can they be observed only in the far northern or southern latitudes? Is there a relationship between the auroras and surface activity on the Sun, called solar flares? Are they related to other physical phenomena observed on Earth? Finally, how can an understanding of the science of the auroras aid in the development of new technologies? Your studies in this chapter will help answer these questions. C H A P T E R 12 Key Concepts In this chapter, you will learn about: magnetic fields moving charges in magnetic and electric fields electromagnetic induction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define electric current as the amount of charge passing a reference point per unit of time describe magnetic interactions in terms of forces and fields compare gravitational, electric, and magnetic fields describe how the work of Oersted and Faraday led to the theory relating electricity to magnetism describe a moving charge as the source of a magnetic field and predict the field’s orientation explain how uniform magnetic and electric fields affect a moving charge describe and explain the interaction between a magnetic field and a moving charge and a conductor explain, quantitatively, the effect of an external magnetic field on a current-carrying conductor describe the effects of moving a conductor in an external magnetic field in terms of moving charges Science, Technology, and Society explain that concepts, models, and theories are often used in interpreting, explaining, and predicting observations explain that technology provides solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and vice versa 580 Unit VI 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 581 12-1 QuickLab 12-1 QuickLab Magnetic Fields in a Bottle Problem What is the shape of a magnetic field? Materials 50 mL of iron filings 450 mL of light cooking oil 1 clear plastic 591-mL pop bottle string 1 cylindrical cow magnet (must be able to fit in the bottle) tape Procedure 1 Pour 50 mL of iron filings into the bottle. 2 Pour cooking oil into the bottle until it is about three-quarters full. 3 Replace the cap on the bottle securely and shake the bottle several times so that the iron filings disperse throughout the oil. Remove the cap. 4 Attach the string to one end of the cow magnet and insert the magnet in the bottle. Make sure the magnet is suspended vertically in the middle of the bottle. Tape the other end of the
string to the top of the bottle. 5 Replace the cap on the bottle and place the bottle on a table to allow the mixture to settle. Observe the pattern produced by the iron filings. Questions 1. In your notebook, draw a diagram of the pattern created by the iron filings. 2. 3. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain your answer. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. 4. From the pattern of the iron filings, is it possible to determine the strength and the direction of the magnetic influence around the magnet? Explain your answer. Think About It 1. Describe a probable cause of the pattern of the iron filings in 12-1 QuickLab. 2. What types of substances produce this influence? 3. What types of objects are affected by this influence? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 581 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 582 12.1 Magnetic Forces and Fields An ancient Greek legend from about 800 BCE describes how the shepherd Magnes, while tending his flock, noticed that pieces of a certain type of rock were attracted to the nails on his shoes and to his metal staff (Figure 12.2). This phenomenon was called magnetism and, as time passed, further studies of the behaviour of this rock revealed several curious effects. For example, a piece of this rock could either attract or repel another similar piece (Figure 12.3). This effect seemed to result from two different magnetic effects, so investigators thought that there must be two different types of “magnetic ends,” or poles, on the rock. This observation led to the law of magnetism, which states: Like magnetic poles repel and unlike poles attract each other. (a) (b) Figure 12.3 A piece of magnetic rock, held near one end of a similar piece of magnetic rock, would attract at one end (a) and repel at the other end (b). In 1269, Pierre de Maricourt was mapping the position of a magnetized needle placed at various positions on the surface of a spherical piece of this rock. He observed that the
directions of the needle formed a pattern that encircled the rock, like meridian lines, and converged at two points on opposite ends of the rock. When this rock was then suspended by a string, the two converging points tended to align along Earth’s north–south axis. This property of the rock earned it the name “lodestone” or “leading stone.” Maricourt called the end pointing northward the north-seeking or north pole and the end pointing southward the south-seeking or south pole. All magnets have both poles. Lodestone, which contains the mineral magnetite (Fe3O4), was later used in the development of compass technology. Figure 12.2 The magnetic effects of certain materials were observed by ancient Greeks as early as 800 BCE info BIT Magnetic poles always exist in pairs. In the 1930s, Paul Dirac (1902–1984) suggested the existence of a particle called a magnetic monopole. To date, all experiments to discover this onepoled particle have failed, but these particles are still under experimental investigation. 582 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 583 Concept Check Figure 12.4 A U-magnet, a circular magnet, and a bar magnet Copy the picture of each magnet in Figure 12.4 into your notebook. Since each magnet must have two poles, label the possible positions of the north and south poles of each magnet. The next big advance in knowledge about magnetism came from the work of William Gilbert. In his book De Magnete, published in 1600, he not only reviewed and criticized past explanations of magnetism but he also presented many important new hypotheses. He compared the orientation of magnetized needles on the surface of a spherical piece of lodestone with the north–south orientation of a compass needle at various locations on Earth’s surface. From this study, he proposed that Earth itself is a lodestone with north and south magnetic poles. Concept Check The north pole of a magnetic compass needle points toward Earth’s magnetic north. What can you conclude about this point on Earth? Gilbert was also intrigued by the forces that magnets could exert on other magnetic objects. If you suspend a magnet on a string and bring another magnet close to one of its poles, the suspended magnet will rotate, even though there is no visible contact between the two magnets. Magnets appeared to have the ability to exert forces that
seemed to originate from the magnetic poles, and they could affect another magnetic object even without contact. The ancient Greeks called this effect “action at a distance.” Recall from chapter 11 that they used the same terminology to describe the effects of electric charges. In attempting to explain the action at a distance caused by a magnet, Gilbert suggested that an invisible “orb of virtue” surrounds a magnet and extends in all directions around it. Other magnetic substances react to a force created by this orb of virtue and move or rotate in response. His orbs of virtue were the beginnings of the idea of “fields” that would revolutionize physics. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 583 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 584 magnetic field: a threedimensional region of influence surrounding a magnet, in which other magnets are affected by magnetic forces Michael Faraday (1791–1867) further developed this concept. He defined a magnetic field as a three-dimensional region of magnetic influence surrounding a magnet, in which other magnets are affected by magnetic forces. The direction of the magnetic field at a given location is defined as the direction in which the north pole of the compass needle points at that location. Some materials, such as iron, act like magnets while in a magnetic field. 12-2 QuickLab 12-2 QuickLab Observing Magnetic Fields Problem How can the magnitude and direction of magnetic fields be observed and analyzed? Questions 1. Describe the cause of the pattern produced by the iron filings. 2. 3. 4. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. Is it possible to determine the strength and direction of the magnetic field surrounding the magnet from the pattern of the iron filings alone? Explain your answer. 5. From your investigation of the effect of a magnetic field on a compass, what appears to be the direction of the magnetic field around a magnet? Materials 1 bar magnet 1 sheet of paper (216 mm 279 mm) 25 mL of iron filings 1 compass Procedure 1 Lay the bar magnet on a table and place the paper over the magnet. Trace the shape of the magnet on the paper and label the poles. 2 Carefully sprinkle the iron filings onto the surface of the paper. 3 Tap the paper lightly to reinforce the alignment of the
iron filings on the sheet. Draw the pattern of the iron filings around the magnet. 4 Clean the iron filings from the paper and replace the paper over the magnet. 5 Place the compass at several positions around the magnet and trace the direction of the compass needle. 584 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 585 Magnetic Fields The magnetic field surrounding a magnet is represented by the symbol B and is measured in teslas (T). A typical bar magnet in the classroom can have a magnetic field of approximately 1 102 T, whereas Earth’s magnetic field is about 5 105 T. The magnetic field is a vector quantity, so it is represented by a vector arrow. In diagrams, the length of the vector arrow represents the magnitude of the field, and the direction of the arrow represents the direction of the field at a point. You can also use compasses to show the direction of the magnetic field at any position surrounding a magnet, as illustrated in Figure 12.5. Figure 12.5 shows that, in general, this direction is from the north to the south pole of the magnet. info BIT Magnetic field lines run parallel to Earth’s surface only at the equator. As they reach the magnetic poles, they gradually dip toward the surface. At the poles, the magnetic field lines point perpendicular to Earth’s surface. Navigators in the far north or south must be aware that the magnetic compasses may be of limited use in those areas Figure 12.5 The direction of a magnetic field is the direction of the force on the north pole of a compass placed in the field. To represent the entire magnetic field surrounding a magnet, it would be necessary to draw arrows at an infinite number of points around the magnet. This is impractical. Instead, you can draw a few magnetic field lines with a single arrow head indicating the direction of the magnetic field. To find the field direction at a given point, move the arrow head along the field line through that point so that it keeps pointing in the direction of the tangent to the field line. The field lines in Figure 12.6 are a map of the magnetic field with the following features: • Outside a magnet, the magnetic field lines point away from the north pole of a magnet and toward the south pole. • The closeness of the lines represents the magnitude of the magnetic field. e LAB For a probeware activity where you use a magnetic field sensor to determine the relationship between
the distance from a magnet and the intensity of the field, go to www.pearsoned.ca/ school/physicssource. S N S N (a) (b) Figure 12.6 (a) The pattern of iron filings surrounding a bar magnet outlines the magnetic field. (b) Magnetic field lines, representing the direction and magnitude of the magnetic field, can replace the iron filings. The number of magnetic field lines that exit a magnetic material is equal to the number of magnetic field lines that enter the magnetic material, forming closed loops. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 585 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 586 PHYSICS INSIGHT Before the adoption of SI units, magnetic fields were sometimes measured in a CGS unit called the gauss (G). You might see this unit in some older books. 1 T 104 G. Table 12.1 shows some examples of magnetic field strengths. Table 12.1 Magnetic Field Strengths Physical system Magnetic field (T) Earth Bar magnet Sunspots High field magnetic resonance imaging device (MRI) Strongest humanmade magnetic field Magnetar (magnetic neutron star) Concept Check 5 105 1 102 1 101 15 40 1 1011 Figure 12.7 shows the patterns produced by iron filings that are influenced by the magnetic fields of one or two magnets. Sketch the magnetic field lines in each case. S N N N N S N S Figure 12.7 Concept Check List at least two similarities and two differences between gravitational, electric, and magnetic fields. 586 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/28/08 10:04 AM Page 587 Cause of Magnetism The force of magnetic repulsion between like poles of magnets is the same force that causes the almost frictionless ride of the Maglev (magnetically levitated) train (Figure 12.8). What is the source of this “magnetic levitation” on the train? info BIT Oersted was among the first to recognize the talent of the writer Hans Christian Andersen and encouraged him when he began writing his now famous fairy tales. (a) (b) Figure 12.8 (a) The force of magnetic repulsion between like poles can cause one magnet to levitate over another. (b) The Maglev train, developed in Japan, floats several centimetres above the guideway, providing a smooth and almost
frictionless ride. Experiments by early investigators revealed many facts about the magnetic fields surrounding magnets and their effects on magnetic objects. However, the actual cause of magnetism eluded scientists until 1820. While demonstrating to students that the current passing through a wire produces heat, Danish professor Hans Christian Oersted (1777–1851) noticed that the needle of a nearby compass deflected each time the circuit was switched on. This experiment led Oersted to the important conclusion that there is a relationship between electricity and magnetism, at a time when electricity and magnetism were considered separate phenomena. He proved that electric current was a cause of magnetism. Following his initial observations, it was later shown that if electric current was in a straight line, the magnetic field formed a circular pattern (Figure 12.9(a)), and if the electric current was circular, the magnetic field was straight within the coil (Figure 12.9(b)). (a) battery N (b) electron flow electron flow S Figure 12.9 (a) A current passing through a straight conducting wire produces a magnetic field, represented by concentric red circular lines around the wire. (b) A current passing through a coil produces a magnetic field, represented by red circular lines, with poles similar to those of a bar magnet. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 587 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 588 PHYSICS INSIGHT The observation of a magnetic field produced by a moving charge depends on the frame of reference of the observer. If you are stationary and the charge moves past you, you observe a magnetic field. However, if you are moving along with the charge, the charge is stationary relative to you, so you do not observe a magnetic field. Left-hand Rules for Magnetic Fields A useful left-hand rule to determine the direction of the magnetic field is the wire-grasp rule described in Figure 12.10. To determine the direction of the magnetic field produced by a moving charge, use the left-hand wire-grasp rule if the moving charge is negative. (If the moving charge is positive, then use the right-hand wire-grasp rule.) direction of magnetic field lines (a) e direction of electron flow left hand conductor magnetic field lines (b) core e direction of electron flow magnetic field line e direction of magnetic field Figure 12.10 Left-hand rule for direction of a magnetic field due to
moving charges: (a) If the conducting wire is straight, then the thumb indicates the direction of the straight current and the cupped fingers indicate the direction of the circular magnetic field. (b) If the current is in a coil of conducting wire, the cupped fingers indicate the circular current and the straight thumb indicates the direction of the straight magnetic field within the coil Using the Wire-grasp Rule 1. Sketch the following diagrams into your notebook. (a) e Indicate the direction of the magnetic field lines and the direction of current in the wire, as required. (b) N S Electromagnets As shown in Figure 12.9(b), current in a circular loop or coil of wire produces a magnetic field like that of a bar magnet. An electromagnet uses a current-carrying coil of wire to generate a magnetic field that is easy to switch on and off. The strength of an electromagnet can be influenced by: • increasing the current through the wire • increasing the number of loops in the coil • increasing the size of the loops in the coil • changing the core of the coil Powerful electromagnets have many industrial uses, such as lifting steel parts, machinery, or scrap iron. Electromagnets are widely used to remotely operate switches or valves. Often, a valve is activated by a metal rod that is drawn into the core of the electromagnet when current electromagnet: a magnet having its magnetic field produced by electric current flowing through a coil of wire 588 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 589 flows through the coil. Such mechanisms, called solenoids, are common in washing machines, dishwashers, furnaces, and industrial machinery. Figure 12.11 shows two applications of electromagnets. solenoid: an electromagnet that operates a mechanical device (a) (b) Figure 12.11 (a) A lifting magnet (b) An appliance solenoid Domain Theory and Magnetization In some atoms, the configuration of the electrons is such that their movement generates a tiny magnetic field. In ferromagnetic materials, such as iron, nickel, and cobalt, the magnetic fields of adjacent atoms can align to reinforce each other, forming small regions, or domains, with intense magnetic fields. Domains generally range from 0.001 mm to 1 mm across, and may contain billions of atoms. The orientations of the magnetic fields of the various
domains are normally random, so their magnetic fields largely balance each other, leaving the material with little or no overall magnetization. However, the size of a domain and the direction of its magnetic field are relatively easy to change. An external magnetic field can cause the domains to align, thus magnetizing the material (Figure 12.12). The small black arrows in Figure 12.12 indicate the orientation of the magnetic field of an individual domain. S N (a) (b) Figure 12.12 (a) When the magnetic fields of atoms in a region line up, they create a magnetic domain in the substance. (b) Aligning the domains produces a magnet. A typical ferromagnetic object has vastly more domains than the diagrams can show. If you hang an iron nail by a string and bring a magnet close to the nail, the nail will rotate toward the magnet, even before they touch. The nail is not a magnet with distinct poles, yet a magnetic attraction exists between it and the magnet. When the magnet is close to the nail, the domains in the nail that are oriented for attraction to the magnet increase in size while the other domains shrink. When the magnet is moved away again, the domains in the nail tend to return to random info BIT Geophysicists theorize that circulating currents of ions in the molten core of Earth produce its magnetic field. ferromagnetic: having magnetic properties like those of iron domain: a region of a material in which the magnetic fields of most of the atoms are aligned e WEB All magnetic substances can be classified as one of the following: • ferromagnetic • antiferromagnetic • ferrimagnetic • paramagnetic • diamagnetic Find out what distinguishes one type of magnetic substance from another. Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 589 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 590 orientations and the nail loses most of its magnetization. This example illustrates induced magnetization. The nail will be much more strongly magnetized if it is stroked with a pole of the magnet. The magnetic fields of many of the domains in the nail will align along the direction of motion of the magnet. This magnetization is strong enough that the nail will remain somewhat magnetized when the magnet is removed. Concept Check A filing cabinet has been in one position for a long time. It is made of fer
romagnetic material, so it can become a permanent magnet. If you hold a compass near the top of the filing cabinet, the compass needle points toward the filing cabinet. If you hold the compass near the bottom of the filing cabinet, the opposite end of the compass points toward the cabinet. Has the cabinet been magnetized by Earth’s magnetic field? Or has the cabinet become magnetized by the magnetic compass? Explain your answer. Magnetism in Nature The effects of magnetism have been known since early civilizations, but the causes of magnetic behaviour are only now being revealed. A modern understanding of magnetic phenomena began with the development of field theory to replace “action at a distance.” The symmetry of nature enabled scientists to use the same field theory to describe the gravitational field surrounding any mass, the electric field surrounding any charge, and the magnetic field surrounding any magnet. Oersted’s investigations, which revealed a relationship between electricity and magnetism, ultimately led to the domain theory to explain a cause of magnetism. As scientists probed deeper into the mysteries of magnetism, many more answers were found. However, the tremendous significance of magnetism has only recently been understood in explaining phenomena and producing technological applications. In the field of biology, for example, researchers have found that certain organisms have ferromagnetic crystals consisting of magnetite in their bodies. Some bacteria use these magnetite crystals to help orient themselves within Earth’s magnetic field. Bees and pigeons have magnetite crystals within their brains to help with navigation. The human brain also has these magnetite crystals, but their function is not clear. It is known that an external magnetic field can disrupt the neural activity in the parietal lobe on one side of the human brain. 590 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 591 Understanding magnetism has also led to important technological advancements. These advancements range from simple applications, such as refrigerator magnets, magnetic stripes on cards, and magnetic audiocassette or VCR tapes, to more complicated applications involving magnetic levitation, such as the Maglev train and magnetic resonance imaging (MRI) machines used as a diagnostic tool in health care. Although much has been achieved, there are still many secrets of magnetism to uncover. THEN, NOW, AND FUTURE Earth’s Magnetic Field William Gilbert’s “Terrella” experiment in the 1500s compared the magnetic field of Earth to that of a bar
magnet. From that time, Earth has been considered to be a huge magnet, with similar magnetic properties to a much smaller, ordinary magnet. This observation was successful in explaining many phenomena. However, care must be taken in comparing the causes of magnetic behaviour in Earth and in a bar magnet. If the cause of magnetism in substances is the motion of charges, scientists are not quite convinced that the motion of charges within Earth’s molten core is responsible for Earth’s magnetism. They know that Earth’s molten core is simply too hot for atoms to remain aligned and exhibit any magnetic properties. Other probable causes of Earth’s magnetic field could be convection currents rising to the cooler surface of Earth, or the motion of charges in the upper ionosphere. The most acceptable and probable cause, though, is the motion of charges in the molten part of Earth, just beneath the crust (Figure 12.13). Whatever the cause of Earth’s magnetic behaviour, it is known that the magnetic field of Earth is not stable. Molten rock within the interior of Earth has no magnetic properties. However, when molten rock rises to the surface, it cools and solidifies, and its domains orient themselves in line with Earth’s magnetic field at the time. When samples of rock from different strata formed throughout geological times are tested, evidence shows that there are times when not only the magnitude of Earth’s magnetic field changed, but also its direction. In the past five million years, more than 20 reversals have occurred, the last one about 780 000 years ago. Coincidentally, modern humans emerged during this time period. One possible effect of a zero magnetic field, during a reversal, would be an increase in the cosmic ray intensity at Earth’s surface. Normally, the magnetic field shields Earth from harmful radiation from space. Fossil evidence indicates that periods of no protective magnetic field have been effective in changing life forms. Evidence that these types of changes could have occurred also comes from heredity studies of fruit flies when exposed to X rays. We cannot know precisely when the next reversal will occur. However, evidence from recent measurements indicates a decrease in the magnitude of Earth’s magnetic field of about 5% in the last 100 years. Based on this evidence, Figure 12.13 This computer model of Earth shows the molten outer core surrounding the inner core (the small circle). The right side shows the molten currents. The left side shows the magnetic field lines that extend outward
through the rest of Earth’s interior. another reversal of Earth’s magnetic field may occur within the next 2000 years. Questions 1. Can the motion of charges in Earth’s core create domains? Explain your answer. 2. What is the most probable cause of Earth’s magnetic behaviour? 3. What evidence is there on Earth that its magnetic field is not stable? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 591 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 592 12.1 Check and Reflect 12.1 Check and Reflect Knowledge 1. What is the law of magnetism? 2. Explain your answers to the following: (a) Does every magnet have a north and a south pole? 10. List at least two differences and two similarities between (a) gravitational and electric fields (b) gravitational and magnetic fields (c) electric and magnetic fields (b) Does every charged object have positive and negative charges? 11. Using the domain theory, explain the following observations: 3. How did William Gilbert determine that Earth was a magnet? 4. What is the most probable cause of magnetism in (a) a bar magnet? (b) Earth? 5. What accidental discovery did Oersted make? 6. What is the shape of the magnetic field (a) around a straight current-carrying conductor? (b) within a coil of conducting wire carrying a current? (a) A magnet attracts an unmagnetized ferromagnetic material. (b) Stroking a nail with a magnet magnetizes the nail. (c) A metal table leg affects a compass. 12. Why does dropping or heating a bar magnet decrease its magnetic properties? 13. Consider a bar magnet and Earth, as shown below. Describe the similarities and the differences of their magnetic fields. S N Earth Applications 7. What would happen to a magnet if you broke it into two pieces? Extensions 8. A negatively charged sphere is approaching you. Describe the magnetic field surrounding the sphere and its direction. What would happen if the sphere were positively charged? 9. A spinning top is charged negatively and is spinning clockwise, as observed from above. Describe the magnetic field created by the spinning top and its direction. 14. Why is it difficult to get an accurate bearing with a magnetic compass near the poles? 15. Do magnetic field lines always run parallel to the surface of Earth? Explain your answer
. 16. If a current-carrying wire is bent into a loop, why is the magnetic field stronger inside the loop than outside? e TEST To check your understanding of magnetic forces and fields, follow the eTest links at www.pearsoned.ca/school/physicssource. 592 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 593 12.2 Moving Charges and Magnetic Fields Near the end of the 1800s, researchers were fascinated by a new technology called the cathode-ray tube (CRT), shown in Figure 12.14. It consisted of a glass tube from which air had been evacuated, and it had a positive plate (anode) at one end and a negative plate (cathode) at the other end. These new tubes used electric fields to accelerate a beam called a cathode ray through a large potential difference. The beam would “light up” the fluorescent screen at the end of the tube. Scientists were unsure whether this beam was a type of electromagnetic radiation (similar to light), a neutral particle, or a charged particle. They initially called it a cathode ray because it appeared to originate from the cathode plate. This technology not only enabled the discovery of the electron in 1897 (see section 15.1), but also led to the later development of many other technologies, including television. Until recently, the image in most TVs was produced by an electron beam striking a fluorescent screen in a CRT. The Motor Effect Deflecting charged particles involves an interaction of two magnetic fields. A charged particle in uniform motion produces a circular magnetic field around it (the wire-grasp rule). Now suppose this charged particle enters an external magnetic field, produced between the faces of two opposite magnetic poles. The interaction of the circular magnetic field of the charge and the external magnetic field produces a magnetic force that acts on the particle to deflect it, as shown in Figure 12.15. This magnetic force is also called the motor effect force (F m) because it causes the rotation of a loop of current-carrying wire. This rotation is fundamental in the operation of an electric motor. Figure 12.14 The image on the fluorescent screen at the left end of this cathode-ray tube shows that the rays originate from the negative terminal at the right end. motor effect force: the deflecting force acting on a charged particle moving in a magnetic field Figure 12.15 (a) The cathode
ray accelerates in a straight line when it is only influenced by the electric field produced between the cathode and anode plates in a vacuum tube connected to a high-voltage source. (b) A cathode ray will deflect as shown when it is also under the influence of an external magnetic field. (a) (b) In Figure 12.16, the straight, horizontal lines represent the external magnetic field of the magnetic poles, and the dashed lines represent the magnetic field surrounding the moving charge (using the left-hand rule). In the “replacement magnet” method of illustration, tiny magnets are drawn along the field lines to reinforce the idea of the direction and the effects of the interaction of the two magnetic fields. The represents Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 593 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 594 PHYSICS INSIGHT represents a direction into the page, like the fletching of an arrow receding from you. • represents a direction out of the page, like the point of an arrow approaching you. In Figure 12.16 represents negative charge movement into the page. negative charge moving into the page. In Figure 12.16, below the moving charge, the external magnetic field and the magnetic field surrounding the charge are in the same direction. Above the moving charge, the two magnetic fields are in opposite directions. Fm causing deflection Fm Fm S N Figure 12.16 The combined magnetic forces due to the two magnetic fields cause the moving charge to deflect (F perpendicular to the direction of the external magnetic field. m) in a direction perpendicular to its direction of motion and Since the external magnetic field is fixed, the combined effect of m) on the the two magnetic fields produces a net magnetic force (F moving charge. As a result, the moving charge deflects upward (toward the top of the page). The deflecting force is always perpendicular to the direction of both the external magnetic field and the motion of the moving charge, as shown in Figure 12.16. This property distinguishes a magnetic field from electric or gravitational fields. Since the direction of the electric force or gravitational force can be parallel to their respective fields, these fields can be used to change the speed of a charged particle. The magnetic force, on the other hand, is always perpendicular to the velocity of the charged particle. A magnetic force can never do any work on a charged particle, nor can it change
the speed or kinetic energy of a charged particle. Since force is not in the direction of the displacement, then there can be no work done on the object. Only the direction of the charged particle’s path may be changed. Left-hand Rule for Deflection Consider a negatively charged particle travelling perpendicular to an external magnetic field. When it enters the region of a uniform magnetic field, it is deflected in a direction perpendicular to both the original direction of charge movement and the direction of the external magnetic field. A useful hand rule to determine the direction of deflection is the left-hand rule shown in Figure 12.17: 594 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 595 • The thumb indicates the direction of the initial charge movement. • The extended fingers indicate the direction of the external magnetic field, from north to south. • The palm faces in the direction of the magnetic force. If the moving charge is positive, use the righthand rule, with the thumb, fingers, and palm indicating the directions of the same quantities as in the left-hand rule. S magnetic field B e N Fm v Figure 12.17 How to use the left-hand rule to determine the deflection of a charged particle Using the Left-hand Rule for Deflection 1. In your notebook, sketch the direction of the unknown variable in each situation. (a) external magnetic field (b) negative charge Fm e motion B Fm 12-3 Inquiry Lab 12-3 Inquiry Lab Using Hand Rules with a Cathode-ray Tube — Demonstration (c) positive charge motion B Fm Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Do the hand rules predict the deflection of a cathode beam in an external magnetic field? Materials and Equipment 1 cathode-ray tube 1 high-voltage source 1 strong bar magnet S N cathode-ray tube high-voltage source Figure 12.18 CAUTION! High voltage. Be very careful around electrical equipment to avoid shocks. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 595 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 596 Procedure Analysis 1 Connect the cathode-ray tube to the high-voltage 1. What types of particles would be attracted from the source, as shown in Figure 12.18. Identify the cath
ode and anode of the CRT. 2 Turn on the current supply. Observe the path of the cathode rays that are produced. 3 Carefully hold the north pole of a bar magnet near one side of the centre of the cathode tube, in the horizontal plane. Note the direction in which the cathode rays are deflected. 4 Repeat the procedure in step 3 with the south pole of the bar magnet. 5 Repeat steps 3 and 4 by holding the magnet on the other side of the cathode tube, in the horizontal plane and then in the vertical plane. Observe the direction of the deflection of the beam in each case. cathode to the anode of the CRT? Does the hand rule applicable for these particles predict all of the deflections that you observed? 2. What types of particles would be attracted from the anode to the cathode of the CRT? Does the hand rule for these particles predict any of the deflections that you observed? 3. Explain how the observed deflections show that cathode rays consist of charged particles. 4. Can you use the hand rules to determine the type of charge carried by cathode rays? Explain why or why not. Concept Check Compare the magnetic force of an external magnetic field on a moving charged particle with: • the gravitational force of Earth on the mass of the charged particle • the electric force due to another nearby charged particle Charged Particle Motion in a Magnetic Field The direction of the initial motion of a charged particle in an external magnetic field determines how the charged particle will deflect. Figure 12.19 shows what can happen to a charged particle as it enters an external magnetic field: (a) If the initial motion of the charged particle is parallel to the external magnetic field, then there is no effect. (b) If the initial motion of the charged particle is perpendicular to the external magnetic field, the charge is deflected in a circular arc. (c) If the initial motion of the charged particle is at an angle to an external magnetic field, the charge deflects in a circular motion that will form a helical path. e SIM Explore the motion of a charged particle in a uniform magnetic field. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 596 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 597 (a) (c) v v v (b) B B (d) B
Fm Fm Fm B Figure 12.19 (a) When the charged particle’s velocity is parallel to the external magnetic field (B), the charged particle’s path is a straight line. (b) The charged particle’s motion is perpendicular to the magnetic field, so the particle is deflected in a circular arc. (c) The charged particle’s motion is at an angle to the magnetic field, so the particle follows a helical path. (d) This side view from the left shows the magnetic force acting as the centripetal force that causes the charge to follow a circular path. Oppositely charged particles deflect in opposite directions in a magnetic field (Figure 12.20). If the magnitude of the external magnetic field is large enough, the field can cause circular motion that remains contained in the magnetic field. In this circular motion, the centripetal force is the magnetic force. Magnetic deflection of charged particles is the underlying principle for useful powerful analytical and research tools such as mass spectrometers and particle accelerators. Unit VIII presents these devices and their applications in science, medicine, and industry. Auroras Tremendous expulsions of magnetic energy from the solar atmosphere, called solar flares, expel streams of charged particles at speeds around 10% of the speed of light (Figure 12.21). When some of these particles strike Earth’s magnetic field, they are deflected by the magnetic force and spiral in a helical path along Earth’s magnetic field lines. These particles enter the atmosphere as they approach Earth’s magnetic poles, and collide with air molecules. These collisions excite the atoms of the air molecules, in a process that will be described in Chapter 15, causing them to emit visible light that we see as the aurora. The process repeats because Earth’s nonuniform magnetic field produces a magnetic force component that causes the charged particles to reverse their direction of motion, travelling to Earth’s opposite pole. The same auroral effect is produced at this pole, and the process continues to repeat as the charged particles oscillate back and forth between the poles, trapped in a type of “magnetic bottle” called the Van Allen belt. N S upward deflection downward deflection Figure 12.20 A magnetic field deflects moving oppositely charged particles in opposite directions, as shown. e WEB Research the formation of the Van Allen belts. Consider the shape of the Van Allen belt on the side of
Earth facing the Sun and on the side of Earth away from the Sun. What is the cause of this difference in shape? Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 597 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 598 Figure 12.21 This composite image shows the cause of the aurora borealis. Streams of high-energy charged particles erupt from the Sun (far left). They are deflected by Earth’s magnetic field toward the poles, creating the bright ring shown in the satellite image of Earth (centre). There they interact with air molecules in the atmosphere to produce the aurora (far right). Calculating the Magnetic Force By studying the different types of deflections, scientists can also explain the complex deflection of charged particles entering a magnetic field at an angle, such as the particles that cause the auroras. The magnitude of the deflecting force (F m ) depends on all of the following: • the magnitude of the moving charge (q) • the magnitude of the perpendicular velocity component (v) ) • the magnitude of the external magnetic field (B The magnitude of the deflecting force can be calculated using this equation: F m qvB where q is the magnitude of moving charge in coulombs (C); v is the component of the speed perpendicular to the magnetic field in metres per second (m/s); and B is the magnitude of the external magnetic field in teslas (T). Example 12.1 describes how to calculate the magnetic force on a charge moving perpendicular to an external magnetic field. When the velocity of the charge is not perpendicular to the magnetic field, you can use trigonometry to find the perpendicular component: v v sin where is the angle between the charge’s velocity, v, and the magnetic field, B. v v v 598 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 599 Example 12.1 An electron is travelling at 3.20 105 m/s perpendicular to an external magnetic field of magnitude 2.20 101 T (Figure 12.22). Determine the magnetic force acting on the electron. S N S N e Figure 12.22 Given q charge on 1 electron 1.60 1019 C 2.20 101 T B v 3.20 105 m/
s Required m) magnetic force (F Analysis and Solution Determine the magnitude of the magnetic deflecting force: F m qvB (1.60 1019 C)(3.20 105 m )(2.20 101 T) s 1.13 1014 N Since the charge is negative, use the left-hand rule for deflection to determine the direction of the magnetic force. • Thumb points in the direction of the charged particle’s movement, into the page. • Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). • Palm points in the direction of the magnetic deflecting force, toward the top of the page. Practice Problems 1. A proton with a charge of 1.60 1019 C is travelling with a speed of 3.50 104 m/s perpendicularly through an external magnetic field of magnitude 4.20 104 T. Determine the magnitude of the magnetic deflecting force on the proton. 2. An ion with a charge of 3.20 1019 C and a speed of 2.30 105 m/s enters an external magnetic field of 2.20 101 T, at an angle of 30, as shown in the figure below. Calculate the magnitude of the magnetic deflecting force on the ion. 30° S N 3. A negatively charged sphere travels from west to east along Earth’s surface at the equator. What is the direction of the magnetic deflecting force on the sphere? Answers 1. 2.35 1018 N 2. 8.10 1015 N 3. Downward toward Earth’s surface Paraphrase The magnetic force is 1.13 1014 N [upward] (toward the top of the page). Often, a charged particle may be influenced by a combination of two fields, such as a magnetic field and a gravitational field, or a magnetic field and an electric field. “Crossed-field” devices are technologies that use both magnetic and electric fields. An example is the magnetron, which produces microwaves in microwave ovens. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 599 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 600 Example 12.2 A carbon ion, with a mass of 2.01 1026 kg and a positive charge of magnitude 1.60 1019 C, enters the region of an external magnetic field of magnitude 6.32 105
T, as shown in Figure 12.23. Find the perpendicular speed at which the magnetic deflecting force will balance the gravitational force such that the carbon ion will travel in a straight line. Practice Problems 1. An electron, with a charge of magnitude 1.60 1019 C and a mass of 9.11 1031 kg, is travelling west along the surface of Earth at the equator. If the magnitude of the magnetic field at this location is 5.00 105 T, what minimum speed must the electron maintain to remain at the same height above Earth’s surface? 2. Ions, with a charge of 1.60 1019 C and a mass of 8.12 1026 kg, travel perpendicularly through a region with an external magnetic field of 0.150 T. If the perpendicular speed of the ions is 8.00 104 m/s, determine (a) the magnitude of the deflecting force on the ion (b) the radius of curvature of the motion of the deflected ion Hint: The magnetic deflecting force is the centripetal force. F F m c v2 m qvB r Answers 1. 1.12 106 m/s 2. (a) 1.92 1015 N (b) 0.271 m S N carbon ion Figure 12.23 Given m 2.01 1026 kg 6.32 105 T B q 1.60 1019 C g 9.81 N/kg Required speed (v) at which the magnitudes of the magnetic force, F m and the gravitational force, F g, are equal, Analysis and Solution The gravitational force on the carbon ion has a magnitude of mg and is directed downward (toward the bottom of F g the page). The magnetic force on the carbon ion has a magnitude of qvB F m the page). and must be directed upward (toward the top of F net F m F g But the magnetic deflecting force and the gravitational force balance (Figure 12.24), so F net 0. Therefore, F m qvB F g mg g v m B q (2.01 1026 kg)9.81 N k g (6.32 105 T)(1.60 1019 C) 1.95 102 m/s Fm Fg Paraphrase The carbon atom will travel in a straight line if its speed is 1.95 102 m/s. Figure 12.24 600 Unit VI Forces and Fields 12-PearsonPhys30-Chap
12 7/24/08 3:35 PM Page 601 In this section, you have studied the deflection of a moving charged particle in a magnetic field. Applying this science, you learned not only the importance of this phenomenon in technologies, such as a television and a magnetron, but also the significance of this phenomenon in protecting Earth from harmful cosmic radiations. The magnetic field of Earth, in deflecting dangerous charged particles from striking Earth’s surface, also produces one of the most beautiful and spectacular natural light shows—the aurora. 12.2 Check and Reflect 12.2 Check and Reflect Knowledge 1. Why is a cathode ray called a cathode ray? 2. What is the difference between a magnetic field vector arrow and a magnetic field line? 3. An electron and a proton, both with the same perpendicular velocity, enter a region with a uniform external magnetic field. What can you state about the deflections of both particles? 4. Describe the key differences in how magnetic and electric fields affect a moving charged particle. Applications 5. A positively charged lithium ion is travelling horizontally along Earth’s surface. Describe the deflection due to the magnetic force if the ion travels (a) south to north (b) east to west (c) upward into the atmosphere 6. A proton with a speed of 2.00 105 m/s enters an external magnetic field of magnitude 0.200 T. Calculate the magnitude of the deflecting force if the proton enters (a) perpendicular to the magnetic field (b) at an angle of 35.0 to the field 7. A 0.020-g metal ball with a charge of 3.0 C is thrown horizontally along Earth’s equator. How fast must the ball be thrown so that it maintains the same height, during its motion tangential to Earth’s surface, if the magnitude of Earth’s magnetic field is 5.0 105 T? 8. An alpha particle, with a charge of 2 1.60 1019 C, is travelling perpendicularly through a magnetic field of magnitude 2.00 102 T at a speed of 1.02 105 m/s. What minimum gravitational force is required to suspend the alpha particle at the same position above Earth’s surface? 9. Electrons in the picture tube of a television are accelerated to a speed of 1.30 106 m/s. As they travel through the tube, they experience a perpendicular magnetic field of magnitude 0.0700
T. What is the radius of deflection of the electrons in the tube? 10. A cosmic ray proton travelling through space at 4.38 106 m/s deflects in a circular arc with a radius of 5.50 106 m. What is the magnitude of the magnetic field at that point in space? Extensions 11. Why are auroras seen only at higher latitudes? e TEST To check your understanding of moving charges and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 601 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 602 12.3 Current-carrying Conductors and Magnetic Fields info BIT The first sensitive meter to measure small currents was developed by Luigi Galvani in the 1700s. While dissecting a frog’s leg, he noticed that an electric current caused the frog’s leg to twitch. He realized that he had accidentally discovered a method of detecting small currents and used this discovery to design the galvanometer. current: the quantity of charge that flows through a wire in a given unit of time ampere: the flow of 1 C of charge past a point in a conductor in 1 s 602 Unit VI Forces and Fields Figure 12.25 A galvanometer and an electric motor, like the one in this lawn mower, apply magnetic fields produced by a flow of charge. Two of the most common applications of magnetic fields acting on moving charged particles are meters (such as ammeters, voltmeters, and galvanometers) and electric motors (Figure 12.25). Although these technologies appear to be different from the technology of the television, the basic operating principle of all these technologies is similar. Recall from earlier science studies that a galvanometer is a device for detecting and measuring small electric currents. How does a galvanometer operate? How is its operation similar to the technologies of the electric motor and television? Electric Current Recall from earlier science courses that electric current is the movement of charged particles. It can be defined more precisely as the quantity of charge that flows through a wire in a given unit of time. The unit for current, the ampere (A), is a measure of the rate of current. The ampere is an SI base unit. A current of 1 A is equivalent to the flow of 1 C of charge past a point in a conductor in 1 s. In
other words, 1 A 1 C/s. For example, the effective value of the current through a 100-W light bulb is about one ampere (1 A) of current. The ampere is named in honour of the French scientist André-Marie Ampère (1775–1836), who is renowned for his analysis of the relationship between current and magnetic force. This equation shows the relationship between current and charge: q I t where I is the current in amperes, q is the magnitude of charge in coulombs, and t is the time elapsed in seconds. 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 603 Example 12.3 Calculate the current in a wire through which 20.0 C of charge passes in 4.00 s. Given q 20.0 C t 4.00 s Required current (I) Analysis and Solution To calculate the current, use the equation I q t.0 C 2 0 4 s 0.0 5.00 C s 5.00 A Paraphrase The current in the conducting wire is 5.00 A. Practice Problems 1. A lightning strike transfers 20.0 C of charge to the ground in 1.00 ms. Calculate the current during this lightning strike. 2. If the current in a household appliance is 5.00 A, calculate the amount of charge that passes through the appliance in 10.0 s. Answers 1. 2.00 104 A 2. 50.0 C Magnetic Force on a Current-carrying Conductor In a CRT picture tube, powerful external magnetic fields are used to deflect moving electrons to produce an image on a screen. To analyze the operation of a galvanometer or electric motor, and to reveal the similarity of their operation to that of a television, consider the movement of electrons as a current in a wire conductor. When there is an electric current in a wire that is perpendicular to an external magnetic field, each electron experiences a magnetic force caused by the interactions of its own magnetic field and the external magnetic field (Figure 12.26). You can observe the effect of this force. The magnetic force causes the electrons to deflect upward. However, the electrons cannot escape the wire, so if the magnetic force on the electrons is great enough, the whole wire will rise upward, opposite to the force of gravity. The magnetic force on a conducting wire is the same as the magnetic m) that you studied in section 12.2. deflecting force on
a moving charge (F S N S N e Figure 12.26 A current of electrons passes through a conducting wire lying perpendicular to an external magnetic field. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 603 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 604 PHYSICS INSIGHT Remember: If the moving charges are negative, use your left hand; and if the moving charges are positive, use your right hand. Left-hand Rule for Magnetic Force To determine the direction of the magnetic force, you can use the lefthand rule, as shown in Figure 12.27: • Your thumb indicates the direction of electron flow in the conductor. • Your extended fingers point in the direction of the external magnetic field. • Your palm indicates the direction of the magnetic deflecting force on the wire. S N S N e Fm v B Figure 12.27 The left-hand rule for determining the direction of magnetic force To calculate the magnitude of the magnetic force for a length of current-carrying conducting wire, use the equation F m IlB where I is the current measured in amperes; l is the length of the wire perpendicular to the magnetic field in metres; B is the magnitude of the external magnetic field in teslas; and F is m the magnitude of the magnetic force in newtons. The Galvanometer In the operation of the galvanometer, a coil of wire is mounted to allow for movement within the strong magnetic field of the permanent magnet (Figure 12.28). The coil turns against a spring with an attached needle pointing to a calibrated scale. When there is a current in the coil, the magnetic forces cause the coil to rotate. The greater the current, the greater the rotation, as registered on the scale by the needle. The galvanometer, which measures very small currents, can be made to measure larger currents (ammeter) by connecting a small resistance in parallel, and to measure larger potential differences (voltmeter) by connecting a large resistance in series. The magnetic force produced on a current-carrying wire can be demonstrated in the 12-4 QuickLab on page 606. N S current restoring spring Figure 12.28 A schematic diagram of a galvanometer reveals all the essential components in its operation. 604 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 605 Example 12.4 An 8.50-cm length
of conducting wire lies perpendicular to an external magnetic field of magnitude 4.20 mT, as shown in Figure 12.29. If there is a negative charge flow of 2.10 A in the conductor, calculate the magnitude and determine the direction of the magnetic force on the wire. S N S N Figure 12.29 Given l 8.50 cm 8.50 102 m 4.20 mT 4.20 103 T B I 2.10 A Practice Problems 1. A 0.500-m length of conducting wire carrying a current of 10.0 A is perpendicular to an external magnetic field of magnitude 0.200 T. Determine the magnitude of the magnetic force on this wire. 2. A thin conducting wire 0.75 m long has a mass of 0.060 kg. What is the minimum current required in the wire to make it “float” in a magnetic field of magnitude 0.15 T? Required magnitude and direction of the magnetic force on the wire (F m) Answers 1. 1.00 N 2. 5.2 A Analysis and Solution Determine the magnitude of the magnetic force: F m IlB (2.10 A)(8.50 102 m)(4.20 103 T) 7.50 104 N Use the left-hand rule to determine the direction of the magnetic force, because the moving charges are negative: • Thumb points in the direction of the charge movement or current, into the page. • Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). • Palm points in the direction of the magnetic force, to the top of the page. Paraphrase The magnetic force is 7.50 104 N [upward] (toward the top of the page). Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 605 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 606 12-4 QuickLab 12-4 QuickLab Demonstration of a Current-carrying Conductor in a Uniform Magnetic Field Problem How does a uniform magnetic field affect a currentcarrying conductor? 2 Carefully increase the current (amperage) from the power supply. 3 Observe any effects on the current-carrying conductor. Materials 1 piece of stiff insulated conducting wire (6–8 cm long) 2 alligator clips 1 U-shaped magnet thread or light string retort stand and clamp variable low-voltage DC power supply with
ammeter Questions 1. Describe any effects on the current-carrying conductor that occurred as the current through the conducting wire increased. 2. Does the hand rule verify the direction of the movement of the conducting wire? Explain which hand rule must be used. Procedure 1 Set up the apparatus as shown in Figure 12.30. 3. What is the effect of an external magnetic field on a current-carrying conductor? 4. Based on what you have just observed, design a lab that would demonstrate the effects of a uniform magnetic field on a current-carrying conductor. string or thread ON OFF retort stand power supply insulated wire N S magnet Figure 12.30 606 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 607 Magnetic Forces Between Two Current-carrying Conductors After Oersted demonstrated that a current-carrying conductor creates a magnetic field around a conductor, the French scientist André-Marie Ampère performed extensive studies to determine the magnitude of the magnetic field at any point surrounding a current-carrying conductor. In addition to his mathematical analysis of magnetic fields, he is also noted for determining that two current-carrying conductors exert magnetic forces on each other. The charged particles in one wire are affected by magnetic forces when placed in the magnetic field of another current-carrying wire. Currents in the same direction attract each other (Figure 12.31(a)), and currents in opposite directions repel each other (Figure 12.31(b)). magnetic field wire x e Fm Fm wire x e magnetic field (a) (b) Fm magnetic field wire x e wire e Fm magnetic field Figure 12.31 From the left-hand rule for magnetic fields, the red dashed arrows indicate the orientation of the magnetic field around each wire. Use the left-hand rule for magnetic force to determine how the wires will move relative to each other. (a) When currents are in the same direction, the wires attract each other. (b) When currents are in opposite directions, the wires repel each other. Through careful experimentation and measurement, Ampère was able to determine that the magnetic force between two current-carrying conductors depends on all of the following: • the length of the conducting wire • the distance between the two conducting wires • the amount of current in each wire The SI unit for current is named in honour of Ampère’s work. This unit, the ampere,
is now defined as the current required in each of two current-carrying wires, 1 m long and separated by 1 m in air, to produce a force of 2 107 N of magnetic attraction or repulsion. As you learned at the beginning of this section, an ampere is equivalent to the flow of 1 C of charge in 1 s. So, 1 A 1 C/s, and 1 C 1 As. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 607 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 608 Concept Check In intricate electrical circuits, two conducting wires carrying currents in opposite directions are usually crossed. What is the purpose of this crossing procedure? 12-5 Design a Lab 12-5 Design a Lab Using the Current Balance to Measure the Magnetic Force Between Two Current-carrying Conducting Wires The Question How can you use a current balance to investigate the factors that influence the magnetic force acting on two current-carrying conducting wires? Design and Conduct Your Investigation Study the operation of the current balance in your laboratory and design an experimental procedure to investigate the factors that determine the magnetic force acting on two current-carrying conducting wires. In your experimental design: • Identify the factors that determine the magnetic force acting on two currentcarrying conductors. • Write an “if/then” hypothesis statement that predicts how changes in the variable affect the magnetic force. • Clearly outline the procedure you will perform to investigate the relationship of each factor on the magnetic force. • Describe what you will measure and how the data will be recorded and analyzed. • Explain how the data will be used to answer the question. As a group, identify and designate tasks. Prepare a report that describes your experimental design and present it to your teacher. After approval, conduct the investigation and answer the question. How well did your results agree with your hypothesis? info BIT Advancements in technology make it possible to construct extremely small electric motors. Today, 1000 of the smallest electric motors could fit in the period at the end of this sentence. commutator: a mechanism for maintaining a properly polarized connection to the moving coil in a motor or generator The Electric Motor The most important application of the effect of an external magnetic field on current-carrying conductors is the electric motor. Figure 12.32 illustrates a simple electric motor that works with a current-carrying wire loop between two magnetic poles. The current is in one direction. Recall from earlier science

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