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We found in Sect. 2.3.1 that the electric field can be written as the gradient of a scalar potential.
$$\mathbf{E} = -\nabla V$$
The question arises: What do the divergence and curl of E,
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$
and $\nabla \times \mathbf{E} = \mathbf{0}$ ,
look like, in terms of ... | {
"Header 1": "2.3.3 ■ Poisson's Equation and Laplace's Equation",
"token_count": 523,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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I defined V in terms of $\mathbf{E}$ (Eq. 2.21). Ordinarily, though, it's $\mathbf{E}$ that we're looking for (if we already knew $\mathbf{E}$ , there wouldn't be much point in calculating V). The idea is that it might be easier to get V first, and then calculate $\mathbf{E}$ by taking the gradient. Typically, t... | {
"Header 1": "2.3.4 ■ The Potential of a Localized Charge Distribution",
"token_count": 1997,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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In terms of *r* and the total charge on the shell, *q* = 4π *R*<sup>2</sup>σ,
$$V(r) = \begin{cases} \frac{1}{4\pi\epsilon_0} \frac{q}{r} & (r \ge R), \\ \frac{1}{4\pi\epsilon_0} \frac{q}{R} & (r \le R). \end{cases}$$
Of course, in this particular case, it was easier to get *V* by using Eq. 2.21 than Eq. 2.30, be... | {
"Header 1": "2.3.4 ■ The Potential of a Localized Charge Distribution",
"token_count": 594,
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In the typical electrostatic problem you are given a source charge distribution $\rho$ , and you want to find the electric field **E** it produces. Unless the symmetry of the problem allows a solution by Gauss's law, it is generally to your advantage to calculate the potential first, as an intermediate step. These are... | {
"Header 1": "2.3.5 ■ Boundary Conditions",
"token_count": 1600,
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Suppose you have a stationary configuration of source charges, and you want to move a test charge *Q* from point **a** to point **b** (Fig. 2.39). *Question:* How much work will you have to do? At any point along the path, the electric force on *Q* is **F** = *Q***E**; the force *you* must exert, in opposition to this ... | {
"Header 1": "**2.4.1 The Work It Takes to Move a Charge**",
"token_count": 556,
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How much work would it take to assemble an entire *collection* of point charges? Imagine bringing in the charges, one by one, from far away (Fig. 2.40). The first charge, $q_1$ , takes *no* work, since there is no field yet to fight against. Now bring in $q_2$ . According to Eq. 2.39, this will cost you $q_2V_1(\mat... | {
"Header 1": "2.4.2 ■ The Energy of a Point Charge Distribution",
"token_count": 1463,
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For a volume charge density ρ, Eq. 2.42 becomes
$$W = \frac{1}{2} \int \rho V \, d\tau. \tag{2.43}$$
(The corresponding integrals for line and surface charges would be λ*V dl* and σ *V da*.) There is a lovely way to rewrite this result, in which ρ and *V* are eliminated in favor of **E**. First use Gauss's law to e... | {
"Header 1": "**2.4.3 The Energy of a Continuous Charge Distribution**",
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Use Eq. 2.45. Inside the sphere, $\mathbf{E} = \mathbf{0}$ ; outside,
$$\mathbf{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{\mathbf{r}}, \quad \text{so} \quad E^2 = \frac{q^2}{(4\pi\epsilon_0)^2 r^4}.$$
Therefore,
$$W_{\text{tot}} = \frac{\epsilon_0}{2(4\pi\epsilon_0)^2} \int_{\text{outside}} \left(\frac{q^... | {
"Header 1": "**2.4.3 The Energy of a Continuous Charge Distribution**",
"Header 2": "**Solution 2**",
"token_count": 469,
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(i) A perplexing "inconsistency." Equation 2.45 clearly implies that the energy of a stationary charge distribution is always *positive*. On the other hand, Eq. 2.42 (from which 2.45 was in fact derived), can be positive or negative. For instance, according to Eq. 2.42, the energy of two equal but opposite charges a di... | {
"Header 1": "2.4.4 ■ Comments on Electrostatic Energy",
"token_count": 1434,
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In an **insulator**, such as glass or rubber, each electron is on a short leash, attached to a particular atom. In a metallic **conductor**, by contrast, one or more electrons per atom are free to roam. (In liquid conductors such as salt water, it is ions that do the moving.) A *perfect* conductor would contain an *unl... | {
"Header 1": "2.5.1 ■ Basic Properties",
"token_count": 1119,
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If you hold a charge +*q* near an uncharged conductor (Fig. 2.44), the two will attract one another. The reason for this is that *q* will pull minus charges over to the near side and repel plus charges to the far side. (Another way to think of it is that the charge moves around in such a way as to kill off the field of... | {
"Header 1": "**2.5.2 Induced Charges**",
"token_count": 2017,
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Because the field inside a conductor is zero, boundary condition 2.33 requires that the field immediately *outside* is
$$\mathbf{E} = \frac{\sigma}{\epsilon_0} \hat{\mathbf{n}},\tag{2.48}$$
consistent with our earlier conclusion that the field is normal to the surface. In terms of potential, Eq. 2.36 yields
$$\si... | {
"Header 1": "**2.5.3 Surface Charge and the Force on a Conductor**",
"token_count": 1122,
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Suppose we have *two* conductors, and we put charge +*Q* on one and −*Q* on the other (Fig. 2.51). Since *V* is constant over a conductor, we can speak unambiguously of the potential difference between them:
$$V = V_{+} - V_{-} = -\int_{(-)}^{(+)} \mathbf{E} \cdot d\mathbf{l}.$$
We don't know how the charge distrib... | {
"Header 1": "**2.5.4 Capacitors**",
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**Problem 2.45** Find the electric field at a height *z* above the center of a square sheet (side *a*) carrying a uniform surface charge σ. Check your result for the limiting cases *a* → ∞ and *z a*.
$$\left[ Answer: (\sigma/2\epsilon_0) \left\{ (4/\pi) \tan^{-1} \sqrt{1 + (a^2/2z^2)} - 1 \right\} \right]$$
**Probl... | {
"Header 1": "**More Problems on Chapter 2**",
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What is the charge density? How do you account for the fact that the field points in a particular direction, when the charge density is uniform? [This is a more subtle problem than it looks, and worthy of careful thought.]
2.5 Conductors 111
**Problem 2.56** All of electrostatics follows from the $1/r^2$ characte... | {
"Header 1": "**More Problems on Chapter 2**",
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"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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#### 3.1.1 ■ Introduction
The primary task of electrostatics is to find the electric field of a given stationary charge distribution. In principle, this purpose is accomplished by Coulomb's law, in the form of Eq. 2.8:
$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\hat{\mathbf{z}}}{v^2} \rho(\mathb... | {
"Header 1": "3.1 ■ LAPLACE'S EQUATION",
"token_count": 696,
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Suppose V depends on only one variable, x. Then Laplace's equation becomes
$$\frac{d^2V}{dx^2} = 0.$$
The general solution is
$$V(x) = mx + b, (3.6)$$
the equation for a straight line. It contains two undetermined constants (m and b), as is appropriate for a second-order (ordinary) differential equation. They a... | {
"Header 1": "3.1.2 ■ Laplace's Equation in One Dimension",
"token_count": 542,
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If *V* depends on two variables, Laplace's equation becomes
$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0.$$
This is no longer an *ordinary* differential equation (that is, one involving ordinary derivatives only); it is a *partial* differential equation. As a consequence, some of the ... | {
"Header 1": "**3.1.3 Laplace's Equation in Two Dimensions**",
"token_count": 1011,
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In three dimensions I can neither provide you with an explicit solution (as in one dimension) nor offer a suggestive physical example to guide your intuition (as I did in two dimensions). Nevertheless, the same two properties remain true, and this time I will sketch a proof.<sup>3</sup>
1. The value of *V* at point *... | {
"Header 1": "**3.1.4 Laplace's Equation in Three Dimensions**",
"token_count": 1065,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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Laplace's equation does not by itself determine *V*; in addition, suitable boundary conditions must be supplied. This raises a delicate question: What are appropriate boundary conditions, sufficient to determine the answer and yet not so strong as to generate inconsistencies? The one-dimensional case is easy, for here ... | {
"Header 1": "**3.1.5 Boundary Conditions and Uniqueness Theorems**",
"token_count": 828,
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The potential on the cavity wall is some constant, *V*<sup>0</sup> (that's item (iv), in Sect. 2.5.1), so the potential inside is a function that satisfies Laplace's equation and has the constant value *V*<sup>0</sup> at the boundary. It doesn't take a genius to think of *one* solution to this problem: *V* = *V*<sup>0<... | {
"Header 1": "**Solution**",
"token_count": 530,
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The *simplest* way to set the boundary conditions for an electrostatic problem is to specify the value of *V* on all surfaces surrounding the region of interest. And this situation often occurs in practice: In the laboratory, we have conductors connected to batteries, which maintain a given potential, or to **ground**,... | {
"Header 1": "**3.1.6 Conductors and the Second Uniqueness Theorem**",
"token_count": 1722,
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Suppose a point charge *q* is held a distance *d* above an infinite grounded conducting plane (Fig. 3.10). *Question:* What is the potential in the region above the plane? It's not just (1/4π0)*q*/r, for *<sup>q</sup>* will induce a certain amount of negative charge on the nearby surface of the conductor; the total pot... | {
"Header 1": "**3.2.1 The Classic Image Problem**",
"token_count": 816,
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Now that we know the potential, it is a straightforward matter to compute the surface charge $\sigma$ induced on the conductor. According to Eq. 2.49,
$$\sigma = -\epsilon_0 \frac{\partial V}{\partial n},$$
where $\partial V/\partial n$ is the normal derivative of V at the surface. In this case the normal direc... | {
"Header 1": "3.2.2 ■ Induced Surface Charge",
"token_count": 553,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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The charge q is attracted toward the plane, because of the negative induced charge. Let's calculate the force of attraction. Since the potential in the vicinity of q is the same as in the analog problem (the one with +q and -q but no conductor), so also is the field and, therefore, the force:
$$\mathbf{F} = -\frac{1}... | {
"Header 1": "3.2.3 ■ Force and Energy",
"token_count": 693,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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The method just described is not limited to a single point charge; *any* stationary charge distribution near a grounded conducting plane can be treated in the same way, by introducing its mirror image—hence the name **method of images**. (Remember that the image charges have the *opposite sign*; this is what guarantees... | {
"Header 1": "3.2.4 ■ Other Image Problems",
"token_count": 1551,
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In this section we shall attack Laplace's equation directly, using the method of **separation of variables**, which is the physicist's favorite tool for solving partial differential equations. The method is applicable in circumstances where the potential (*V*) or the charge density (σ) is specified on the boundaries of... | {
"Header 1": "**3.3 SEPARATION OF VARIABLES**",
"token_count": 293,
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The configuration is independent of z, so this is really a *two*-dimensional problem. In mathematical terms, we must solve Laplace's equation,
$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0, (3.20)$$
subject to the boundary conditions
(i)
$$V = 0$$
when $y = 0$ ,
(ii) $V = 0$ when ... | {
"Header 1": "**3.3 SEPARATION OF VARIABLES**",
"Header 2": "**Solution**",
"token_count": 2013,
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3.31 by $\sin(n'\pi y/a)$ (where n' is a positive integer), and integrate from 0 to a:
$$\sum_{n=1}^{\infty} C_n \int_0^a \sin(n\pi y/a) \sin(n'\pi y/a) \, dy = \int_0^a V_0(y) \sin(n'\pi y/a) \, dy. \quad (3.32)$$
You can work out the integral on the left for yourself; the answer is
$$\int_0^a \sin(n\pi y/a) \... | {
"Header 1": "**3.3 SEPARATION OF VARIABLES**",
"Header 2": "**Solution**",
"token_count": 2003,
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3.35:
$$C_n \cosh(n\pi b/a) = \begin{cases} 0, & \text{if } n \text{ is even} \\ \frac{4V_0}{n\pi}, & \text{if } n \text{ is odd} \end{cases}$$
Conclusion: The potential in this case is given by
$$V(x, y) = \frac{4V_0}{\pi} \sum_{n=1,3,5,...} \frac{1}{n} \frac{\cosh(n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a).$$
(... | {
"Header 1": "**3.3 SEPARATION OF VARIABLES**",
"Header 2": "**Solution**",
"token_count": 2019,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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**Problem 3.15** A rectangular pipe, running parallel to the z-axis (from $-\infty$ to $+\infty$ ), has three grounded metal sides, at y=0, y=a, and x=0. The fourth side, at x=b, is maintained at a specified potential $V_0(y)$ .
- (a) Develop a general formula for the potential inside the pipe.
- (b) Find the p... | {
"Header 1": "**3.3 SEPARATION OF VARIABLES**",
"Header 2": "**Solution**",
"token_count": 268,
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In the examples considered so far, Cartesian coordinates were clearly appropriate, since the boundaries were planes. For round objects, spherical coordinates are more natural. In the spherical system, Laplace's equation reads:
$$\frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial V}{\partial r} \right... | {
"Header 1": "3.3.2 ■ Spherical Coordinates",
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In this case, *Bl* = 0 for all *l*—otherwise the potential would blow up at the origin. Thus,
$$V(r,\theta) = \sum_{l=0}^{\infty} A_l r^l P_l(\cos \theta).$$
(3.66)
<sup>13</sup>In rare cases where the *z* axis is excluded, these "other solutions" do have to be considered.
At r = R this must match the specified f... | {
"Header 1": "**Solution**",
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Conclusion:
$$V(r,\theta) = -E_0 \left( r - \frac{R^3}{r^2} \right) \cos \theta. \tag{3.76}$$
The first term $(-E_0 r \cos \theta)$ is due to the external field; the contribution attributable to the induced charge is
$$E_0 \frac{R^3}{r^2} \cos \theta$$
.
If you want to know the induced charge density, it ca... | {
"Header 1": "**Solution**",
"token_count": 2015,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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2.25, you found the potential on the axis of a uniformly charged disk:
$$V(r,0) = \frac{\sigma}{2\epsilon_0} \left( \sqrt{r^2 + R^2} - r \right).$$
- (a) Use this, together with the fact that *Pl*(1) = 1, to evaluate the first three terms in the expansion (Eq. 3.72) for the potential of the disk at points *off* the... | {
"Header 1": "**Solution**",
"token_count": 533,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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Let <sup>r</sup><sup>−</sup> be the distance from <sup>−</sup>*<sup>q</sup>* and <sup>r</sup><sup>+</sup> the distance from <sup>+</sup>*<sup>q</sup>* (Fig. 3.26). Then
$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{\imath_+} - \frac{q}{\imath_-} \right),$$
and (from the law of cosines)
$$n_{\pm}^2 = ... | {
"Header 1": "**Solution**",
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Ordinarily, the multipole expansion is dominated (at large r) by the monopole term:
$$V_{\text{mon}}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r},\tag{3.97}$$
where $Q = \int \rho d\tau$ is the total charge of the configuration. This is just what we expect for the approximate potential at large distances fr... | {
"Header 1": "3.4.2 ■ The Monopole and Dipole Terms",
"token_count": 1594,
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I mentioned earlier that a point charge at the origin constitutes a "pure" monopole. If it is *not* at the origin, it's no longer a pure monopole. For instance, the charge in Fig. 3.32 has a dipole moment $\mathbf{p} = qd\hat{\mathbf{y}}$ , and a corresponding dipole term in its potential. The monopole potential $(1/... | {
"Header 1": "3.4.3 ■ Origin of Coordinates in Multipole Expansions",
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So far we have worked only with *potentials*. Now I would like to calculate the electric *field* of a (perfect) dipole. If we choose coordinates so that **p** is at the origin and points in the z direction (Fig. 3.36), then the potential at r, $\theta$ is (Eq. 3.99):
$$V_{\text{dip}}(r,\theta) = \frac{\hat{\mathbf{... | {
"Header 1": "3.4.3 ■ Origin of Coordinates in Multipole Expansions",
"Header 2": "3.4.4 ■ The Electric Field of a Dipole",
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**Problem 3.37** In Section 3.1.4, I proved that the electrostatic potential at any point P in a charge-free region is equal to its average value over any spherical surface (radius R) centered at P. Here's an alternative argument that does not rely on Coulomb's law, only on Laplace's equation. We might as well set the ... | {
"Header 1": "More Problems on Chapter 3",
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• **Problem 3.47** Show that the average field inside a sphere of radius *R*, due to all the charge within the sphere, is
$$\mathbf{E}_{\text{ave}} = -\frac{1}{4\pi\epsilon_0} \frac{\mathbf{p}}{R^3},\tag{3.105}$$
where **p** is the total dipole moment. There are several ways to prove this delightfully simple resu... | {
"Header 1": "More Problems on Chapter 3",
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[*Hint:* for distribution 1, use the actual situation; for distribution 2, remove q, and set one of the conductors at potential $V_0$ .]
- (a) Both plates of a parallel-plate capacitor are grounded, and a point charge q is placed between them at a distance x from plate 1. The plate separation is d. Find the induced ... | {
"Header 1": "More Problems on Chapter 3",
"token_count": 1895,
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In this chapter, we shall study electric fields in matter. Matter, of course, comes in many varieties—solids, liquids, gases, metals, woods, glasses—and these substances do not all respond in the same way to electrostatic fields. Nevertheless, *most* everyday objects belong (at least, in good approximation) to one of t... | {
"Header 1": "**4.1.1 Dielectrics**",
"token_count": 282,
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What happens to a neutral atom when it is placed in an electric field **E**? Your first guess might well be: "Absolutely nothing—since the atom is not charged, the field has no effect on it." But that is incorrect. Although the atom as a whole is electrically neutral, there *is* a positively charged core (the nucleus) ... | {
"Header 1": "**4.1.2 Induced Dipoles**",
"token_count": 1882,
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The neutral atom discussed in Sect. 4.1.2 had no dipole moment to start with— $\mathbf{p}$ was *induced* by the applied field. Some molecules have built-in, permanent dipole moments. In the water molecule, for example, the electrons tend to cluster around the oxygen atom (Fig. 4.4), and since the molecule is bent at ... | {
"Header 1": "**4.1.3** ■ Alignment of Polar Molecules",
"token_count": 1642,
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In the previous two sections, we have considered the effect of an external electric field on an individual atom or molecule. We are now in a position to answer (qualitatively) the original question: What happens to a piece of dielectric material when it is placed in an electric field? If the substance consists of neutr... | {
"Header 1": "**4.1.4 Polarization**",
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Suppose we have a piece of polarized material—that is, an object containing a lot of microscopic dipoles lined up. The dipole moment per unit volume **P** is given. *Question:* What is the field produced by this object (not the field that may have *caused* the polarization, but the field the polarization *itself* cause... | {
"Header 1": "**4.2.1** ■ Bound Charges",
"token_count": 1973,
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In the last section we found that the field of a polarized object is identical to the field that would be produced by a certain distribution of "bound charges," $\sigma_b$ and $\rho_b$ . But this conclusion emerged in the course of abstract manipulations on the integral in Eq. 4.9, and left us with no clue as to the... | {
"Header 1": "**4.2.2** ■ Physical Interpretation of Bound Charges",
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I have been sloppy about the distinction between "pure" dipoles and "physical" dipoles. In developing the theory of bound charges, I assumed we were working with the pure kind—indeed, I started with Eq. 4.8, the formula for the potential of a perfect dipole. And yet, an actual polarized dielectric consists of *physical... | {
"Header 1": "**4.2.3 The Field Inside a Dielectric**<sup>4</sup>",
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This is a particularly useful way to express Gauss's law, in the context of dielectrics, because *it makes reference only to free charges,* and free charge is the stuff we control. Bound charge comes along for the ride: when we put the free charge in place, a certain polarization automatically ensues, by the mechanisms... | {
"Header 1": "**4.2.3 The Field Inside a Dielectric**<sup>4</sup>",
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Drawing a cylindrical Gaussian surface, of radius *s* and length *L*, and applying Eq. 4.23, we find
$$D(2\pi sL) = \lambda L.$$
Therefore,
$$\mathbf{D} = \frac{\lambda}{2\pi s} \hat{\mathbf{s}}.\tag{4.24}$$
Notice that this formula holds both within the insulation and outside it. In the latter region, **P** = ... | {
"Header 1": "**Solution**",
"token_count": 980,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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Equation 4.22 looks just like Gauss's law, only the *total* charge density $\rho$ is replaced by the *free* charge density $\rho_f$ , and **D** is substituted for $\epsilon_0 \mathbf{E}$ . For this reason, you may be tempted to conclude that **D** is "just like" **E** (apart from the factor $\epsilon_0$ ), except ... | {
"Header 1": "**4.3.2** ■ A Deceptive Parallel",
"token_count": 1254,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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In Sects. 4.2 and 4.3 we did not commit ourselves as to the *cause* of $\mathbf{P}$ ; we dealt only with the *effects* of polarization. From the qualitative discussion of Sect. 4.1, though, we know that the polarization of a dielectric ordinarily results from an electric field, which lines up the atomic or molecular d... | {
"Header 1": "4.4.1 ■ Susceptibility, Permittivity, Dielectric Constant",
"token_count": 1422,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
To compute *V*, we need to know **E**; to find **E**, we might first try to locate the bound charge; we could get the bound charge from **P**, but we can't calculate **P** unless we already know **E** (Eq. 4.30). We seem to be in a bind. What we *do* know is the *free* charge *Q*, and fortunately the arrangement is sph... | {
"Header 1": "**Solution**",
"token_count": 2018,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
A *crystal* is generally easier to polarize in some directions than in others,<sup>12</sup> and in this case Eq. 4.30 is replaced by the general linear relation
$$P_{x} = \epsilon_{0}(\chi_{e_{xx}}E_{x} + \chi_{e_{xy}}E_{y} + \chi_{e_{xz}}E_{z}) P_{y} = \epsilon_{0}(\chi_{e_{yx}}E_{x... | {
"Header 1": "**Solution**",
"token_count": 954,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In a (homogeneous isotropic) linear dielectric, the bound charge density $(\rho_b)$ is proportional to the free charge density $(\rho_f)$ :<sup>13</sup>
$$\rho_b = -\nabla \cdot \mathbf{P} = -\nabla \cdot \left(\epsilon_0 \frac{\chi_e}{\epsilon} \mathbf{D}\right) = -\left(\frac{\chi_e}{1 + \chi_e}\right) \rho_f. \... | {
"Header 1": "**Solution**",
"Header 2": "**4.4.2** ■ Boundary Value Problems with Linear Dielectrics",
"token_count": 516,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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This is reminiscent of Ex. 3.8, in which an uncharged *conducting* sphere was introduced into a uniform field. In that case, the field of the induced charge canceled **E**<sup>0</sup> within the sphere; in a *dielectric,* the cancellation (from the bound charge) is incomplete.
Our problem is to solve Laplace's equati... | {
"Header 1": "**Solution**",
"token_count": 1985,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Meanwhile, a charge $(q + q_b)$ at (0, 0, d) yields the potential
$$V = \frac{1}{4\pi\epsilon_0} \left[ \frac{q + q_b}{\sqrt{x^2 + y^2 + (z - d)^2}} \right],\tag{4.53}$$
for the region z < 0. Taken together, Eqs. 4.52 and 4.53 constitute a function that satisfies Poisson's equation with a point charge q at (0, 0,... | {
"Header 1": "**Solution**",
"token_count": 1052,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
It takes work to charge up a capacitor (Eq. 2.55):
$$W = \frac{1}{2}CV^2.$$
If the capacitor is filled with linear dielectric, its capacitance exceeds the vacuum value by a factor of the dielectric constant,
$$C = \epsilon_r C_{\text{vac}},$$
as we found in Ex. 4.6. Evidently the work necessary to charge a diel... | {
"Header 1": "**4.4.3 Energy in Dielectric Systems**",
"token_count": 1545,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
From Gauss's law (in the form of Eq. 4.23), the displacement is
$$\mathbf{D}(r) = \begin{cases} \frac{\rho_f}{3} \mathbf{r} & (r < R), \\ \frac{\rho_f}{3} \frac{R^3}{r^2} \hat{\mathbf{r}} & (r > R). \end{cases}$$
So the electric field is
$$\mathbf{E}(r) = \begin{cases} \frac{\rho_f}{3\epsilon_0 \epsilon_r} \mathb... | {
"Header 1": "**Solution**",
"token_count": 2039,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
If it bothers you that the force is taken to be proportional to the separation, look again at Example 4.1.
*friction,* for instance, then *W*spring can be made as large as you like, by assembling the charges in such a way that the spring is obliged to expand and contract many times before reaching its final state. In... | {
"Header 1": "**Solution**",
"token_count": 279,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Just as a conductor is attracted into an electric field (Eq. 2.51), so too is a dielectric—and for essentially the same reason: the bound charge tends to accumulate near the free charge of the opposite sign. But the calculation of forces on dielectrics can be surprisingly tricky. Consider, for example, the case of a sl... | {
"Header 1": "**4.4.4 Forces on Dielectrics**",
"token_count": 1442,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
- (a) For the configuration in Prob. 4.5, calculate the *force* on **p**<sup>2</sup> due to **p**1, and the force on **p**<sup>1</sup> due to **p**2. Are the answers consistent with Newton's third law?
- (b) Find the total torque on **p**<sup>2</sup> *with respect to the center of* **p**1, and compare it with the torqu... | {
"Header 1": "**Problem 4.29**",
"token_count": 1998,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
*Question:* What is the relation between the atomic polarizability $\alpha$ and the susceptibility $\chi_e$ ?
Since **P** (the dipole moment per unit volume) is **p** (the dipole moment per atom) times N (the number of atoms per unit volume), $\mathbf{P} = N\mathbf{p} = N\alpha\mathbf{E}$ , one's first inclinatio... | {
"Header 1": "**Problem 4.29**",
"token_count": 1211,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Remember the basic problem of classical electrodynamics: We have a collection of charges *q*1, *q*2, *q*3, ... (the "source" charges), and we want to calculate the force they exert on some other charge *Q* (the "test" charge). (See Fig. 5.1.) According to the principle of superposition, it is sufficient to find the for... | {
"Header 1": "**5.1.1 Magnetic Fields**",
"token_count": 908,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In fact, this combination of directions is just right for a cross product: the magnetic force on a charge *Q*, moving with velocity **v** in a magnetic field **B**, is<sup>2</sup>
$$\mathbf{F}_{\text{mag}} = Q(\mathbf{v} \times \mathbf{B}). \tag{5.1}$$
This is known as the **Lorentz force law**. <sup>3</sup> In the... | {
"Header 1": "**5.1.2 Magnetic Forces**",
"token_count": 1992,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
For if Q moves an amount $d\mathbf{l} = \mathbf{v} dt$ , the work done is
$$dW_{\text{mag}} = \mathbf{F}_{\text{mag}} \cdot d\mathbf{l} = Q(\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt = 0. \tag{5.11}$$
This follows because $(\mathbf{v} \times \mathbf{B})$ is perpendicular to $\mathbf{v}$ , so $(\mathbf{v... | {
"Header 1": "Magnetic forces do no work.",
"token_count": 733,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The **current** in a wire is the *charge per unit time* passing a given point. By definition, negative charges moving to the left count the same as positive ones to the right. This conveniently reflects the *physical* fact that almost all phenomena involving moving charges depend on the *product* of charge and velocity... | {
"Header 1": "**5.1.3 Currents**",
"token_count": 929,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
First of all, the current must circulate clockwise, in order for (**I** × **B**) in the horizontal segment to point upward. The force is
$$F_{\text{mag}} = IBa,$$
where *a* is the width of the loop. (The magnetic forces on the two vertical segments cancel.) For *F*mag to balance the weight (*mg*), we must therefore... | {
"Header 1": "**Solution**",
"token_count": 1615,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The area (perpendicular to the flow) is π*a*2, so
$$J = \frac{I}{\pi a^2}.$$
This was trivial because the current density was uniform.
(b) Suppose the current density in the wire is proportional to the distance from the axis,
$$J = ks$$
(for some constant *k*). Find the total current in the wire.
![](_page_... | {
"Header 1": "**Solution**",
"token_count": 1264,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
*Stationary charges* produce electric fields that are constant in time; hence the term **electrostatics**. <sup>8</sup> *Steady currents* produce magnetic fields that are constant in time; the theory of steady currents is called **magnetostatics**.
Stationary charges $\Rightarrow$ constant electric fields: electros... | {
"Header 1": "5.2.1 ■ Steady Currents",
"token_count": 654,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The magnetic field of a steady line current is given by the **Biot-Savart law**:
$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{I} \times \hat{\mathbf{i}}}{r^2} dl' = \frac{\mu_0}{4\pi} I \int \frac{d\mathbf{l}' \times \hat{\mathbf{i}}}{r^2}.$$
(5.34)
The integration is along the current path, in ... | {
"Header 1": "5.2.2 ■ The Magnetic Field of a Steady Current",
"token_count": 2057,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
<sup>11</sup> As I mentioned earlier, a point charge does not constitute a steady current, and the Biot-Savart law, which only holds for steady currents, does *not* correctly determine its field.
The superposition principle applies to magnetic fields just as it does to electric fields: if you have a *collection* of s... | {
"Header 1": "5.2.2 ■ The Magnetic Field of a Steady Current",
"token_count": 793,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The magnetic field of an infinite straight wire is shown in Fig. 5.27 (the current is coming *out* of the page). At a glance, it is clear that this field has a nonzero curl (something you'll never see in an *electrostatic* field); let's calculate it.
According to Eq. 5.38, the integral of $\bf B$ around a circular ... | {
"Header 1": "**5.3.1** ■ Straight-Line Currents",
"token_count": 973,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The Biot-Savart law for the general case of a volume current reads
$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}') \times \hat{\mathbf{i}}}{v^2} d\tau'. \tag{5.47}$$
This formula gives the magnetic field at a point $\mathbf{r} = (x, y, z)$ in terms of an integral over the current ... | {
"Header 1": "5.3.2 ■ The Divergence and Curl of B",
"token_count": 1804,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The equation for the curl of **B**.
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J},\tag{5.56}$$
is called **Ampère's law** (in differential form). It can be converted to integral form by the usual device of applying one of the fundamental theorems—in this case Stokes' theorem:
$$\int (\nabla \times \mathbf{B}) \cd... | {
"Header 1": "5.3.3 ■ Ampère's Law",
"token_count": 615,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
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We know the direction of **B** is "circumferential," circling around the wire as indicated by the right-hand rule. By symmetry, the magnitude of **B** is constant around an Amperian loop of radius *s*, centered on the wire. So Ampère's law gives
$$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B2\pi s = \mu_0 I_{\... | {
"Header 1": "**Solution**",
"token_count": 2043,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In that case, it follows that the *magnetic field of the toroid is circumferential at all points, both inside and outside the coil*.
**FIGURE 5.38**
*Proof.* According to the Biot-Savart law, the field at **r** due to the current element at **r** is
$$d\mathbf{B} = \frac{\mu_0}{4\p... | {
"Header 1": "**Solution**",
"token_count": 1502,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The divergence and curl of the *electrostatic* field are
$$\begin{cases} \nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0} \, \rho, & \text{(Gauss's law);} \\ \\ \nabla \times \mathbf{E} = \mathbf{0}, & \text{(no name).} \end{cases}$$
These are **Maxwell's equations** for electrostatics. Together with the boundary con... | {
"Header 1": "**5.3.4 Comparison of Magnetostatics and Electrostatics**",
"token_count": 1464,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Just as **∇** × **E** = **0** permitted us to introduce a scalar potential (*V*) in electrostatics,
$$\mathbf{E} = -\nabla V,$$
so **∇** · **B** = 0 invites the introduction of a *vector* potential **A** in magnetostatics:
$$\mathbf{B} = \nabla \times \mathbf{A}.\tag{5.61}$$
The former is authorized by Theorem ... | {
"Header 1": "**5.4.1 The Vector Potential**",
"token_count": 1852,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
It might seem natural to set the polar axis along ω, but in fact the integration is easier if we let **r** lie on the *z* axis, so that ω is tilted at an angle ψ. We may as well orient the *x* axis so that ω lies in the *x z* plane, as shown in Fig. 5.46. According to Eq. 5.66,
![](_p... | {
"Header 1": "**Solution**",
"token_count": 2015,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Thus
$$\mathbf{A} = \frac{\mu_0 nI}{2} \frac{R^2}{s} \hat{\boldsymbol{\phi}}, \quad \text{for } s \ge R. \tag{5.73}$$
If you have any doubts about this answer, *check* it: Does $\nabla \times \mathbf{A} = \mathbf{B}$ ? Does $\nabla \cdot \mathbf{A} = 0$ ? If so, we're done.
Typically, the direction of $\bf A$ ... | {
"Header 1": "**Solution**",
"token_count": 2050,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
\tag{5.74}$$
As for the tangential components, an Amperian loop running perpendicular to the current (Fig. 5.50) yields
$$\oint \mathbf{B} \cdot d\mathbf{l} = \left( B_{\text{above}}^{\parallel} - B_{\text{below}}^{\parallel} \right) l = \mu_0 I_{\text{enc}} = \mu_0 K l,$$
or
$$B_{\text{above}}^{\parallel} - B_... | {
"Header 1": "**Solution**",
"token_count": 785,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
If you want an approximate formula for the vector potential of a localized current distribution, valid at distant points, a multipole expansion is in order. Remember: the idea of a multipole expansion is to write the potential in the form of a power series in 1/r, where r is the distance to the point in question (Fig. ... | {
"Header 1": "**5.4.3** ■ Multipole Expansion of the Vector Potential",
"token_count": 1967,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
3.37.)
Problem 5.34 Show that the magnetic field of a dipole can be written in coordinatefree form:
$$\mathbf{B}_{\text{dip}}(\mathbf{r}) = \frac{\mu_0}{4\pi} \frac{1}{r^3} \left[ 3(\mathbf{m} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}} - \mathbf{m} \right]. \tag{5.89}$$
**Problem 5.35** A circular loop of wire, wit... | {
"Header 1": "**5.4.3** ■ Multipole Expansion of the Vector Potential",
"token_count": 563,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
**Problem 5.39** Analyze the motion of a particle (charge q, mass m) in the magnetic field of a long straight wire carrying a steady current I.
- (a) Is its kinetic energy conserved?
- (b) Find the force on the particle, in cylindrical coordinates, with I along the z axis.
- (c) Obtain the equations of motion.
- (d) ... | {
"Header 1": "**More Problems on Chapter 5**",
"token_count": 2024,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Determine d such that $\partial^2 B/\partial z^2 = 0$ at the midpoint, and find the resulting magnetic field at the center. [Answer: $8\mu_0 I/5\sqrt{5}R$ ]
**Problem 5.48** Use Eq. 5.41 to obtain the magnetic field on the axis of the rotating disk in Prob. 5.37(a). Show that the dipole field (Eq. 5.88), with the ... | {
"Header 1": "**More Problems on Chapter 5**",
"token_count": 2062,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Does (ii) automatically satisfy $\nabla \cdot \mathbf{A} = 0$ ? [Answer: $(\mu_0 I/2\pi s)$ $(z\,\hat{\mathbf{s}} s\,\hat{\mathbf{z}})$ ]
#### Problem 5.54
- (a) Construct the scalar potential $U(\mathbf{r})$ for a "pure" magnetic dipole **m**.
- (b) Construct a scalar potential for the spinning spherical she... | {
"Header 1": "**More Problems on Chapter 5**",
"token_count": 2029,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
If you ask the average person what "magnetism" is, you will probably be told about refrigerator decorations, compass needles, and the North Pole—none of which has any obvious connection with moving charges or current-carrying wires. Yet all magnetic phenomena are due to electric charges in motion, and in fact, if you c... | {
"Header 1": "**6.1.1 Diamagnets, Paramagnets, Ferromagnets**",
"token_count": 339,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
A magnetic dipole experiences a torque in a magnetic field, just as an electric dipole does in an electric field. Let's calculate the torque on a rectangular current loop in a uniform field **B**. (Since any current loop could be built up from infinitesimal rectangles, with all the "internal" sides canceling, as indica... | {
"Header 1": "**6.1.2 Torques and Forces on Magnetic Dipoles**",
"token_count": 2065,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Electrons not only *spin*; they also *revolve* around the nucleus—for simplicity, let's assume the orbit is a circle of radius R (Fig. 6.9). Although technically this orbital motion does not constitute a steady current, in practice the period $T = 2\pi R/v$ is so short that unless you blink awfully fast, it's going t... | {
"Header 1": "6.1.3 ■ Effect of a Magnetic Field on Atomic Orbits",
"token_count": 1288,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In the presence of a magnetic field, matter becomes *magnetized*; that is, upon microscopic examination, it will be found to contain many tiny dipoles, with a net alignment along some direction. We have discussed two mechanisms that account for this magnetic polarization: (1) paramagnetism (the dipoles associated with ... | {
"Header 1": "**6.1.4 Magnetization**",
"token_count": 866,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
#### 6.2.1 ■ Bound Currents
Suppose we have a piece of magnetized material; the magnetic dipole moment per unit volume, **M**, is given. What field does this object produce? Well, the vector potential of a single dipole **m** is given by Eq. 5.85:
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \frac{\mathbf{m} \time... | {
"Header 1": "**6.1.4 Magnetization**",
"Header 3": "6.2 ■ THE FIELD OF A MAGNETIZED OBJECT",
"token_count": 2058,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In the last section, we found that the field of a magnetized object is identical to the field that would be produced by a certain distribution of "bound" currents, **J***<sup>b</sup>* and **K***b*. I want to show you how these bound currents arise physically. This will be a heuristic argument—the rigorous derivation ha... | {
"Header 1": "**6.2.2 Physical Interpretation of Bound Currents**",
"token_count": 935,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Like the electric field, the actual *microscopic* magnetic field inside matter fluctuates wildly from point to point and instant to instant. When we speak of "the" magnetic field in matter, we mean the *macroscopic* field: the average over regions large enough to contain many atoms. (The magnetization **M** is "smoothe... | {
"Header 1": "**6.2.3** ■ The Magnetic Field Inside Matter",
"token_count": 212,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In Sect. 6.2, we found that the effect of magnetization is to establish bound currents $\mathbf{J}_b = \nabla \times \mathbf{M}$ within the material and $\mathbf{K}_b = \mathbf{M} \times \hat{\mathbf{n}}$ on the surface. The field due to magnetization of the medium is just the field produced by these bound currents... | {
"Header 1": "6.3.1 ■ Ampère's Law in Magnetized Materials",
"token_count": 867,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Copper is weakly diamagnetic, so the dipoles will line up opposite to the field. This results in a bound current running antiparallel to *I*, within the wire, and parallel to *I* along the surface (Fig. 6.20). Just how great these bound currents will
**F... | {
"Header 1": "**Solution**",
"token_count": 1426,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
Equation 6.19 looks just like Ampère's original law (Eq. 5.56), except that the *total* current is replaced by the *free* current, and **B** is replaced by $\mu_0$ **H**. As in the case of **D**, however, I must warn you against reading too much into this correspondence. It does *not* say that $\mu_0$ **H** is "just ... | {
"Header 1": "6.3.2 ■ A Deceptive Parallel",
"token_count": 595,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
The magnetostatic boundary conditions of Sect. 5.4.2 can be rewritten in terms of **H** and the *free* current. From Eq. 6.23 it follows that
$$H_{\text{above}}^{\perp} - H_{\text{below}}^{\perp} = -(M_{\text{above}}^{\perp} - M_{\text{below}}^{\perp}), \tag{6.24}$$
while Eq. 6.19 says
$$\mathbf{H}_{\text{above}}... | {
"Header 1": "**6.3.3** ■ Boundary Conditions",
"token_count": 658,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
In paramagnetic and diamagnetic materials, the magnetization is sustained by the field; when $\bf B$ is removed, $\bf M$ disappears. In fact, for most substances the magnetization is *proportional* to the field, provided the field is not too strong. For
notational consistency with the electrical case (Eq. 4.30), ... | {
"Header 1": "**6.3.3** ■ Boundary Conditions",
"Header 3": "6.4.1 ■ Magnetic Susceptibility and Permeability",
"token_count": 1158,
"source_pdf": "datasets/websources/Physics_v1/Physics/griffiths_4ed.pdf"
} |
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