Electromagnetism 2

In this model the flux of E through a surface S, E == LE . da, (2.11 ) is a measure ofthe "number offield lines" passing through S. I put this in quotes because of course we can only draw a representative sample ofthe field lines-the total number would be infinite. But for a given sampling rate the flux is proportional to the number of lines drawn, because the field strength, remember, is proportional to the density of field lines (the number per unit area), and hence E . da is proportional to the number of lines passing through the infinitesimal area da. (The dot product picks out the component of da along the direction ofE, as indicated in Fig. 2.15. It is only the area in the plane perpendicular to E that we have in mind when we say that the density of field lines is the number per unit area.) This suggests that the flux through any closed surface is a measure of the total charge inside. For the field lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside (Fig. 2.l6a). On the other hand, a charge outside the surface will contribute nothing to the total flux, since its field lines pass in one side and out the other (Fig. 2.16b). This is the essence of Gauss's law. Now let's make it quantitative.

vIn this model the flux of E through a surface S, E == LE . da, (2.11 ) is a measure ofthe "number offield lines" passing through S. I put this in quotes because of course we can only draw a representative sample ofthe field lines-the total number would be infinite. But for a given sampling rate the flux is proportional to the number of lines drawn, because the field strength, remember, is proportional to the density of field lines (the number per unit area), and hence E . da is proportional to the number of lines passing through the infinitesimal area da. (The dot product picks out the component of da along the direction ofE, as indicated in Fig. 2.15. It is only the area in the plane perpendicular to E that we have in mind when we say that the density of field lines is the number per unit area.) This suggests that the flux through any closed surface is a measure of the total charge inside. For the field lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside (Fig. 2.l6a). On the other hand, a charge outside the surface will contribute nothing to the total flux, since its field lines pass in one side and out the other (Fig. 2.16b). This is the essence of Gauss's law. Now let's make it quantitative.

(ii) p = 0 inside a conductor. This follows from Gauss's law: V . E = plEa. IfE = 0, so also is p. There is still charge around, but exactly as much plus charge as minus, so the net charge density in the interior is zero. (iii) Any net charge resides on the surface. That's the only other place it can be. (iv) A conductor is an equipotential. For if a and b are any two points within (or at the surface of) a given conductor, V(b) - Yea) = - J~b E· dl = 0, and hence Yea) = V(b).

(iii) The reference point O. There is an essential ambiguity in the definition of potential, since the choice of reference point 0 was arbitrary. Changing reference points amounts to adding a constant K to the potential: V'(r) = - r E. dl = - r O E· dl- rr E· dl = K + VCr), 101 101 10 where K is the line integral of E from the old reference point 0 to the new one 0/. Of course, adding a constant to V will not affect the potential difference between two points: V/(b) - V/(a) = V(b) - Yea), since the K's cancel out. (Actually, it was already clear from Eq. 2.22 that the potential difference is independent of 0, because it can be written as the line integral of E from a to b, with no reference to 0.) Nor does the ambiguity affect the gradient of V: VV/ = VV, since the derivative of a constant is zero. That's why all such V's, differing only in their choice of reference point, correspond to the same field E. Evidently potential as such carries no real physical significance, for at any given point we can adjust its value at will by a suitable relocation of O. In this sense it is rather like altitude: IfI ask you how high Denver is, you will probably tell me its height above sea level, because that is a convenient and traditional reference point. But we could as well agree to measure altitude above Washington D.C., or Greenwich, or wherever. That would add (or, rather, subtract) a fixed amount from all our sea-level readings, but it wouldn't change anything about the real world. The only quantity of intrinsic interest is the difference in altitude between two points, and that is the same whatever your reference level. Having said this, however, there is a "natural" spot to use for 0 in electrostaticsanalogous to sea level for altitude-and that is a point infinitely far from the charge. Ordinarily, then, we "set the zero of potential at infinity." (Since V (0) = 0, choosing a reference point is equivalent to selecting a place where V is to be zero.) But I must warn you that there is one special circumstance in which this convention fails: when the charge distribution itself extends to infinity. The symptom of trouble, in such cases, is that the potential blows up. For instance, the field of a uniformly charged plane is (aj2Eo)fi., as we found in Ex. 2.4; if we naively put 0 = 00, then the potential at height z above the plane becomes i z I I V(z) = - -adz = --a(z- (0). 00 2Eo 2Eo The remedy is simply to choose some other reference point (in this problem you might use the origin). Notice that the difficulty occurs only in textbook problems; in "real life" there is no such thing as a charge distribution that goes on forever, and we can always use infinity as our reference point.

(iv) Potential obeys the superposition principle. The original superposition principle of electrodynamics pertains to the force on a test charge Q. It says that the total force on Q is the vector sum of the forces attributable to the source charges individually: Dividing through by Q, we find that the electric field, too, obeys the superposition principle: Integrating from the common reference point to r, it follows that the potential also satisfies such a principle: v = VI + V2 + ... That is, the potential at any given point is the sum of the potentials due to all the source charges separately. Only this time it is an ordinary sum, not a vector sum, which makes it a lot easier to work with. (v) Units of Potential. In our units, force is measured in newtons and charge in coulombs, so electric fields are in newtons per coulomb. Accordingly, potential is measured in newton-meters per coulomb orjoules per coulomb. A joule per coulomb is called a volt

(v) E is perpendicular to the surface, just outside a conductor. Otherwise, as in (i), charge will immediately flow around the surface utiti! it kills off the tangential component (Fig. 2.43). (Perpendicular to the surface, charge cannot flow, of course, since it is confined to the conducting object.) I think it is strange that the charge on a conductor flows to the surface. Because of their mutual repulsion, the charges naturally spread out as much as possible, but for all of them to go to the surface seems like a waste ofthe interior space. Surely we could do better, from the point of view of making each charge as far as possible from its neighbors, to sprinkle some of them throughout the volume... Well, it simply is not so. You do best to put all the charge on the surface, and this is true regardless of the size or shape of the conductor. 8 The problem can also be phrased in terms of energy. Like any other free dynamical system, the charge on a conductor will seek the configuration that minimizes its potential energy. What property (iii) asserts is that the electrostatic energy of a solid object (with specified shape and total charge) is a minimum when that charge is spread over the surface. For instance, the energy of a sphere is (1/8lfEo)(q2/ R) ifthe charge is uniformly distributed over the surface, as we found in Ex. 2.8, but it is greater, (3 /20lf EO)(q2/ R), if the charge is uniformly distributed throughout the volume (Prob. 2.32). 2.5.2 Induced Charges 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 wiiI 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 cancel off the field of q for points inside the conductor, where the total field must be zero.) Since the negative induced charge is closer to q, there is a net force of attraction. (In Chapter 3 we shall calculate this force explicitly, for the case of a spherical conductor.)

(~12 is the distance between ql and q2 once they are in position). Now bring in q3; this requires work q3 VI,2(r3), where VI,2 is the potential due to charges ql and q2, namely, (1/4JrEo)(ql/~13 + q2/~23). Thus W3 = I (ql q2) --q3 - + - . 4JrEo ~13 ~23 Similarly, the extra work to bring in q4 will be The total work necessary to assemble the first four charges, then, is w= _1_ (qlq2 + qlq3 + qlq4 + q2q3 + q2q4 + q3q4) . 4JrEo ~12 ~13 ~14 ~23 ~24 ~34 You see the general rule: Take the product of each pair of charges, divide by their separation distance, and add it all up: 1 n n q'q' W- _"''''_1) - 4JrEo LL ... i=1 j~l I) j>i (2.40) (2.41) The stipulation j > i isjust to remind you not to count the same pair twice. A nicer way to accomplish the same purpose is intentionally to count each pair twice, and then divide by 2: 1 n n q'q' W- -"''''~ 8JrEo LL ~i' 1=1 j~l ) Hi (we must still avoid i = j, of course). Notice that in this form the answer plainly does not depend on the orderin which you assemble the charges, since every pair occurs in the sum. Let me next pull out the factor qi: 1 n (n 1 qj) W-- q' --- - 2 1 ?; 4JrEO~ij . j,pi The term in parentheses is the potential at point ri (the position of qi) due to all the other charges-all ofthem, now, not just the ones that were present at some stage in the buildingup process. Thus, 1 n W ="2 LqiV(ri)' i=1 (2.42) That's how much work it takes to assemble a configuration of point charges; it's also the amount of work you'd get back out if you dismantled the system. In the meantime, it 2.4. WORK AND ENERGY INELECTROSTATICS 93 represents energy stored in the configuration ("potential" energy, if you like, though for obvious reasons I prefer to avoid that word in this context)

2.1.3 The Electric Field Ifwe have several point charges qI, qz, ... , qn, at distances!z-j, !z-z, ... ,!z-n from Q, the total force on Q is evidently or where !F= QE,I (2.3) (2.4) 1 ~qiA E(r) == -- L...J 2'4 i. 4rrEo . j!z-' 1= 1 E is called the electric field ofthe source charges. Notice that it is a function of position (r), because the separation vectors 4i depend on the location ofthe field point P (Fig. 2.3). But it makes no reference to the test charge Q. The electric field is a vector quantity that varies

2.1.4 Continuous Charge Distributions Our definition ofthe electric field (Eq. 2.4), assumes that the source ofthe field is a collection of discrete point charges qj. If, instead, the charge is distributed continuously over some region, the sum becomes an integral (Fig. 2.5a): 1 f I , E(r) = -- 24dq. 47fEo 1- (2.5) 62 CHAPTER 2. ELECTROSTATICS If the charge is spread out along a line (Fig. 2.5b), with charge-per-unit-Iength A, then dq = Adl' (where dl' is an element of length along the line); if the charge is smeared out over a surface (Fig. 2.5c), with charge-per-unit-area (J, then dq = (J da' (where da' is an element of area on the surface); and if the charge fills a volume (Fig. 2.5d), with charge-per-unit-volume p, then dq = p dr' (where dr' is an element of volume): dq Adl' (J da' pdr'. Thus the electric field of a line charge is I f A(r') E(r) = -- --z-,,-dl'; 47TEO !z. P for a surface charge, I f (J(r') A , E(r) = -- --2-Iz,da; 47TEO !z. S and for a volume charge, I f per') A E(r) = -- --2-Iz,dr'. 47TEo !z. V (2.6) (2.7) (2.8) Equation 2.8 itself is often referred to as "Coulomb's law," because it is such a short step from the original (2.1), and because a volume charge is in a sense the most general and realistic case. Please note carefully the meaning of Iz, in these formulas. Originally, in Eq. 2.4, Iz,i stood for the vectorfrom the source charge qi to the field pointr. Correspondingly, in Eqs. 2.5-2.8, Iz, is the vector from dq (therefore from dl', da', or dr') to the field point r.z

2.3.3 Poisson's Equation and Laplace's Equation 83 We found in Sect. 2.3.1 that the electric field can be written as the gradient of a scalar potential. E = -VV. The question arises: What do the fundamental equations for E, P V·E= - EO and v x E = 0, look like, in terms of V? Well, V .E = V .(-VV) = - V2V, so, apart from that persisting minus sign, the divergence of E is the Laplacian of V. Gauss's law then says that (2.24) This is known as Poisson's equation. In regions where there is no charge, so that p = 0, Poisson's equation reduces to Laplace's equation, (2.25) We'll explore these equations more fully in Chapter 3. So much for Gauss's law. What about the curl law? This says that V x E = V x (-VV) must equal zero. But that's no condition on V-curl of gradient is always zero. Of course, we used the curl law to show that E could be expressed as the gradient of a scalar, so it's not really surprising that this works out: V x E = 0 permits E = - VV; in return, E = - VV guarantees V x E = O. It takes only one differential equation (Poisson's) to determine V, because V is a scalar; for E we needed two, the divergence and the curl. 2.3.4 The Potential of a Localized Charge Distribution I defined V in terms ofE (Eq. 2.21). Ordinarily, though, it's E that we're looking for (if we already knew 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 E by taking the gradient. Typically, then, we know where the charge is (that is, we know p), and we want to find V. Now, Poisson's equation relates V and p, but unfortunately it's "the wrong way around": it would give us p, if we knew V, whereas we want V, knowing p. What we must do, then, is "invert" Poisson's equation. That's the program for this section, although I shaH do it by roundabout means, beginning, as always, with a point charge at the origin.

2.3.5 Summary; Electrostatic Boundary Conditions In the typical electrostatic problem you are given a sourct:; charge distribution p, and you want to find the electric field E it produces. Unless the symmetry of the problem admits a solution by Gauss's law, it is generally to your advanta~e to calculate the potential first, as an intermediate step. These, then, are the three fundamen~al quantities of electrostatics: p, E, and V. We have, in the course of our discussion, derived all six formulas interrelating them. These equations are neatly summarized in Fig. 2.35. We began with just two experimentalobservations: (1) the principle of superposition-a broad general rule applying to all electromagnetic forces, and (2) Coulomb's law-the fundamental law of electrostatics. From these, all else followed.

2.4.3 The Energy of a Continuous Charge Distribution For a volume charge density p, Eq. 2.42 becomes (2.43) (The corresponding integrals for line and surface charges would be f AV dl and f a V da, respectively.) There is a lovely way to rewrite this result, in which p and V are eliminated in favor of E. First use Gauss's law to express p in terms of E: p = EO V . E, so W = f (V . E)V dr. Now use integration by parts (Eq. 1.59) to transfer the derivative from E to V: But VV = -E, so (2.44) 94 CHAPTER 2. ELECTROSTATICS But what volume is this we're integrating over? Let's go back to the formula we started with, Eq. 2.43. From its derivation, it is clear that we should integrate over the region where the charge is located. But actually, any larger volume would do just as well: The "extra" territory we throw in will contribute nothing to the integral anyway, since p = 0 out there. With this in mind, let's return to Eq. 2.44. What happens here, as we enlarge the volume beyond the minimum necessary to trap all the charge? Well, the integral of E 2 can only increase (the integrand being positive); evidently the surface integral must decrease correspondingly to leave the sum intact. In fact, at large distances from the charge, E goes like IIr 2 and V like IIr, while the surface area grows like r 2 . Roughly speaking, then, the surface integral goes down like llr. Please understand that Eq. 2.44 gives you the correct energy W, whatever volume you use (as long as it encloses all the charge), but the contribution from the volume integral goes up, and that of the surface integral goes down, as you take larger and larger volumes. In particular, why not integrate over all space? Then the surface integral goes to zero, and we are left with W = f E 2 dr.

2.4.4 Comments on Electrostatic Energy (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 2.42, the energy of two equal but opposite charges a distance!z- apart would be -(lj4JTEo)(q2 j!z-). What's gone wrong? Which equation is correct? The answer is that both equations are correct, but they pertain to slightly different situations. Equation 2.42 does not take into account the work necessary to make the point charges in the first place; we started with point charges and simply found the work required to bring them together. This is wise policy, since Eq. 2.45 indicates that the energy of a point charge is in fact infinite: w = EO 2/ (q:) (r 2 sir18drd8d¢) = L roo ~dr = 00. 2(4JTEo) r SJTEo 10 r Equation 2.45 is more complete, in the sense that it tells you the total energy stored in a charge configuration, but Eq. 2.42 is more appropriate when you're dealing with point charges, because we prefer (for good reason!) to leave out that portion of the total energy that is attributable to the fabrication of the point charges themselves. In practice, after all, the point charges (electrons, say) are given to us ready-made; all we do is move them around. Since we did not put them together, and we cannot take them apart, it is immaterial how much work the process would involve. (Still, the infinite energy of a point charge is a recurring source of embarrassment for electromagnetic theory, afflicting the quantum version as well as the classical. We shall return to the problem in Chapter 11.) Now, you may wonder where the inconsistency crept into an apparently water-tight derivation. The "flaw" lies between Eqs. 2.42 and 2.43: In the former, V (ri) represents the potential due to all the other charges but not qi, whereas in the latter, V (r) is the full potential. For a continuous distribution there is no distinction, since the amount of charge right at the point r is vanishingly small, and its contribution to the potential is zero. 96 CHAPTER 2. ELECTROSTATICS (2.46) (ii) Where is the energy stored? Equations 2.43 and 2.45 offer two different ways of calculating the same thing. The first is an integral over the charge distribution; the second is an integral over the field. These can involve completely different regions. For instance, in the case of the spherical shell (Ex. 2.8) the charge is confined to the surface, whereas the electric field is present everywhere outside this surface. Where is the energy, then? Is it stored in the field, as Eq. 2.45 seems to suggest, or is it stored in the charge, as Eq. 2.43 implies? At the present level, this is simply an unanswerable question: I can tell you what the total energy is, and I can provide you with several different ways to compute it, but it is unnecessary to worry about where the energy is located. In the context of radiation theory (Chapter II) it is useful (and in General Relativity it is essential) to regard the energy as being stored in the field, with a density EO 2 • 2 E = energy per umt volume. But in electrostatics one could just as well say it is stored in the charge, with a density pV . The difference is purely a matter of bookkeeping. (iii) The superposition principle. Because electrostatic energy is quadratic in the fields, it does not obey a superposition principle. The energy of a compound system is not the sum of the energies of its parts considered separately-there are also "cross terms"; / E2 dT = feEl +E2)2dT 2 EO f 2 2 (El +E2 + 2El·E2)dT WI + W2 + EO / E] . E2 dT. For example, if you double the charge everywhere, you quadruple the total energy

2.5.3 Surface Charge and the Force on a Conductor Because the field inside a conductor is zero, boundary condition 2.33 requires that the field immediately outside is U A E = - n, (2.48) EO consistent with our earlier conclusion that the field is normal to the surface. In terms of potential, Eq. 2.36 yields av u = -EO -. (2.49) an These equations enable you to calculate the surface charge on a conductor, if you can determine E or V; we shall use them frequently in the next chapter. In the presence of an electric field, a surface charge will, naturally, experience a force; the force per unit area, f, is u E. But there's a problem here, for the electric field is discontinuous at a surface charge, so which value are we supposed to use: Eabove, Ebelow, or something in between? The answer is that we should use the average of the two: I f = uEaverage = 2:U(Eabove + Ebelow). (2.50) Why the average? The reason is very simple, though the telling makes it sound complicated: Let's focus our attention on a small patch of surface surrounding the point in question (Fig. 2.50). Make it tiny enough so it is essentially flat and the surface charge on it is essentially constant. The total field consists of two parts-that attributable to the patch itself, and that due to everything else (other regions of the surface, as well as any external sources that may be present): E = Epatch + Eother. Now, the patch cannot exert a force on itself, any more than you can lift yourself by standing in a basket and pulling up on the handles. The force on the patch, then, is due exclusively to Eother, and this suffers no discontinuity (if we removed the patch, the field in the "hole" would be perfectly smooth). The discontinuity is due entirely to the charge on the patch,

2.5.4 Capacitors 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_ = - E . dl. (-) We don't know how the charge distributes itself over the two conductors, and calculating the field would be a mess, if their shapes are complicated, but this much we do know: E is proportional to Q. For E is given by Coulomb's law: E= --1 / -,Ii PAdT 4JTEo !z.2 '

5 Conductors 2.5.1 Basic Properties In an inSUlator, such as glass or rubber, each electron is attached to a particular atom. In a metallic conductor, by contrast, one or more electrons per atom are free to roam about at will through the material. (In liquid conductors such as salt water it is ions that do the moving.) A perfect conductor would be a material containing an unlimited supply of completely free 2.5. CONDUCTORS 97 charges. In real life there are no perfect conductors, but many substances come amazingly close. From this definition the basic electrostatic properties ofideal conductors immediately follow: (i) E =0 inside a conductor. Why? Because if there were any field, those free charges would move, and it wouldn't be electrostatics any more. Well ... that's hardly a satisfactory explanation; maybe all it proves is that you can't have electrostatics when conductors are present. We had better examine what happens when you put a conductor into an external electric field Eo (Fig. 2.42). Initially, this will drive any free positive charges to the right, and negative ones to the left. (In practice it's only the negative charges-electrons-that do the moving, but when they depart the right side is left with a net positive charge-the stationary nuclei-so it doesn't really matter which charges move; the effect is the same.) When they come to the edge of the material, the charges pile up: plus on the right side, minus on the left. Now, these induced charges produce a field of their own, El, which, as you can see from the figure, is in the opposite direction to Eo. That's the crucial point, for it means that the field of the induced charges tends to cancel offthe originalfield. Charge will continue to flow until this cancellation is complete, and the resultant field inside the conductor is precisely zeroJ The whole process is practically instantaneous.

Although the direct u~e of Gauss's law to compute electric fields is limited to cases of spherical, cylindrical, and planar symmetry, we can put together combinations of objects possessing such symmetry, even though the arrangement as a whole is not symmetrical. For example, invoking the principle of superposition, we could find the field in the vicinity of two uniformly charged parallel cylinders, or a sphere near an infinite charged plane.

An element of surface area on this sphere is R2 sin e' de' d¢', so f R2 sin e' de' d¢' a J R2 +z2 - 2Rz cos e' 2 i Jr sine' 2rrR a de' o JR2+z2_2Rzcose' 2rrR2 a (IhJR2 +z2 - 2Rzcose,)[ 2rr;a (JR2 +z2 +2Rz - J R2 +z2 - 2Rz ) -z- 2rrRa [ y'(R + z)2 - y~-] (R - z)2 . At this stage we must be very careful to take the positive root. For points outside the sphere, z is greaterthan R, and hence J(R - z)2 = z-R; for points inside the sphere, J(R - z)2 = R-z. Thus, Ra R2a -[(R +z) - (z - R)] = -, outside; 2Eoz EOZ Ra Ra --[(R +z) - (R - z)] = -, inside. 2Eoz EO In terms of the total charge on the shell, q = 4rr R2a, V(z) = (l/4rrEO)(q/z) (or, in general, VCr) = (l/4rrEO)(q/r» for points outside the sphere, and (l/4rrEO)(q/R) for points inside. Of course, in this particular case, it was easier to get V by using 2.21 than 2.30, because Gauss's law gave us E with so little effort. But if you compare Ex. 2.7 with Prob. 2.7, you will appreciate the power of the potential formulation.

Because V x E = 0, the line integral of E around any closed loop is zero (that follows from Stokes' theorem). Because :f E . dl = 0, the line integral of E from point a to point b is the same for all paths (otherwise you could go out along path (i) and return along path (ii)-Fig. 2.3Q-and obtain :f E . dl ;;f. 0). Because the line integral is independent of path, we can define a function4 IVCr) == - f: E· dl.l (2.21) Here 0 is some standard reference point on which we have agreed beforehand; V then depends only on the point r. It is called the electric potential. Evidently, the potential difference between two points a and b is V(b) - Yea) -f: E . dl + f: E . dl -f: E . dl - 1° E . dl = -lb E . dl. (2.22) Now, the fundamental theorem for gradients states that V(b) - Yea) = l\vV). dl, so 1b(V V) . dl = -lb E . dl. Since, finally, this is true for any points a and b, the integrands must be equal: IE=-VV·I (2.23) Equation 2.23 is the differential version of Eq. 2.21; it says that the electric field is the gradient of a scalar potential, which is what we set out to prove.

By the way, when I speak of the field, charge, or potential "inside" a conductor, I mean in the "meat" of the conductor; if there is some cavity in the conductor, and within that cavity there is some charge, then the field in the cavity will not be zero. But in a remarkable way the cavity and its contents are electrically isolated from the outside world by the surrounding conductor (Fig. 2.45). No external fields penetrate the conductor; they are canceled at the outer surface by the induced charge there. Similarly, the field due to charges within the cavity is killed off, for all exterior points, by the induced charge on the inner surface. (However, the compensating charge left over on the outer surface of the conductor effectively "communicates" the presence of q to the outside world, as we shall see in Ex. 2.9.) Incidentally, the total charge induced on the cavity wall is equal and opposite to the charge inside, for if we surround the cavity with a Gaussian surface, all points of which are in the conductor (Fig. 2.45), :f E . da = 0, and hence (by Gauss's law) the net enclosed charge must be zero. But Qenc = q + q induced, so q induced = -q.

Divergence and Curl of Electrostatic Fields 2.2.1 Field Lines, Flux, and Gauss's Law 65 (2.10) In principle, we are done with the subject of electrostatics. Equation 2.8 tells us how to compute the field of a charge distribution, and Eq. 2.3 tells us what the force on a charge Q placed in this field will be. Unfortunately, as you may have discovered in working Prob. 2.7, the integrals involved in computing E can be formidable, even for reasonably simple charge distributions. Much of the rest of electrostatics is devoted to assembling a bag of tools and tricks for avoiding these integrals. It all begins with the divergence and curl of E. I shall calculate the divergence of E directly from Eq. 2.8, in Sect. 2.2.2, but first I want to show you a more qualitative, and perhaps more illuminating, intuitive approach. Let's begin with the simplest possible case: a single point charge q, situated at the origin: 1 q A E(r) = ---r. 4nEo r 2 To get a "feel" for this field, I might sketch a few representative vectors, as in Fig. 2.12a. Because the field falls off like 1/r 2, the vectors get shorter as you go farther away from the origin; they always point radially outward. But there is a nicer way to represent this field, and that's to connect up the arrows, to form field lines (Fig. 2.12b). You might think that I have thereby thrown away information about the strength of the field, which was contained in the length of the arrows. But actually I have not. The magnitude of the field is indicated by the density of the field lines: it's strong near the center where the field lines are close together, and weak farther out, where they are relatively far apart. In truth, the field-line diagram is deceptive, when I draw it on a two-dimensional surface, for the density oflines passing through a circle of radius r is the total number divided by the circumference (nj2nr), which goes like (ljr), not Ojr2 ). But if you imagine the model in three dimensions (a pincushion with needles sticking out in all directions), then the density of lines is the total number divided by the area of the sphere (n j4nr 2), which does go like (I jr2 ).

Electric Potential 2.3.1 Introduction to Potential The electric field E is not just any old vector function; it is a very special kind of vector function, one whose curl is always zero. E = yx, for example, could not possibly be an electrostatic field; no set of charges, regardless of their sizes and positions, could ever produce such a field. In this section we're going to exploit this special property of electric fields to reduce a vector problem (finding E) down to a much simpler scalar problem. The first theorem in Sect. 1.6.2 asserts that any vector whose curl is zero is equal to the gradient of some scalar. What I'm going to do now amounts to a proof of that claim, in the context of electrostatics.

Gauss's law is always true, but it is not always useful. If p had not been uniform (or, at any rate, not spherically symmetrical), or ifI had chosen some other shape for my Gaussian surface, it would still have been true that the flux of E is O/EO)q, but I would not have been certain that E was in the saine direction as da and constant in magnitude over the surface, and without that I could not puUIEI out of the integral. Symmetry is crucial to this application of Gauss's law. As far as I know, there are only three kinds of symmetry that work: 1. Spherical symmetry. Make your Gaussian surface a concentric sphere. 2. Cylindrical symmetry. Make your Gaussian surface a coaxial cylinder (Fig. 2.19). 3. Plane symmetry. Use a Gaussian "pillbox," which straddles the surface (Fig. 2.20). Although (2) and (3) technically require infinitely long cylinders, and planes extending to infinity in all directions, we shall often use them to get approximate answers for "long" cylinders or "large" plane surfaces, at points far from the edges.

Howeve~, the gradient of V inherits the discontinuity in E; since E implies that I A VVabove - VVbelow = --an, EO or, more conveniently, aVabove aVbelow I ---- --- = --a an an EO ' where -VV, Eq. 2.33 (2.35) (2.36) av A - = VV·n (2.37) an denotes the normal derivative of V (that is, the rate of change in the direction perpendicular to the surface). Please note that these boundary conditions relate the fields and potentialsjust above and just below the surface. For example, the derivatives in Eq. 2.36 are the limiting values as we approach the surface from either side.

If a cavity surrounded by conducting material is itself empty of charge, then the field within the cavity is zero. For any field line would have to begin and end on the cavity wall, going from a plus charge to a minus charge (Fig. 2.47). Letting that field line be part of a closed loop, the rest of which is entirely inside the conductor (where E = 0), the integral Figure 2.47 2.5. CONDUCTORS 101 :f E· dl is distinctly positive, in violation ofEq. 2.19. It folIows that E = 0 within an empty cavity, and there is in fact no charge on the surface of the cavity. (This is why you are relatively safe inside a metal car during a thunderstorm-you may get cooked, if lightning strikes, but you wilI not be electrocuted. The same principle applies to the placement of sensitive apparatus inside a grounded Faraday cage, to shield out stray electric fields. In practice, the enclosure doesn't even have to be solid conductor-chicken wire wilI often suffice.)

In the case of a point charge q at the origin, the flux of E through a sphere of radius r is 1. E. da = f _1_ (;1'). (r 2 sin8d8d¢r) = ~q. j 4nEo r EO (2.12) Notice that the radius of the sphere cancels out, for while the surface area goes up as r 2 , the field goes down as 1I r 2, and so the product is constant. In terms ofthe field-line picture, this makes good sense, since the same number offield lines passes through any sphere centered at the origin, regardless of its size. In fact, it didn't have to be a sphere-any closed surface, whatever its shape, would trap the same number of field lines. Evidently the .flux through any suiface enclosing the charge is q lEO. Now suppose that instead of a single charge at the origin, we have a bunch of charges scattered about. According to the principle of superposition, the total field is the (vector) sum of all the individual fields: n The flux through a surface that encloses them all, then, is f E . da = t (f Ei . da) = t (E~ qi) . 1=1 1=1 For any closed surface, then, 1. E· da = ~Qenc, j EO S (2.13) where Qenc is the total charge enclosed within the surface. This is the quantitative statement of Gauss's law. Although it contains no information that was not already present in Coulomb's law and the principle of superposition, it is of almost magical power, as you will see in Sect. 2.2.3. Notice that it all hinges on the 1I r 2 character of Coulomb's law; without that the crucial cancellation of the r's in Eq. 2.12 would not take place, and the total flux of E would depend on the surface chosen, not merely on the total charge enclosed. Other 1I r 2 forces (I am thinking particularly of Newton's law of universal gravitation) will obey "Gauss's laws" of their own, and the applications we develop here carryover directly. As it stands, Gauss's law is an integral equation, but we can readily turn it into a differential one, by applying the divergence theorem: f E . da = f (V . E) dr. S V Rewriting Qenc in terms of the charge density p, we have Qenc = f pdr.

Let's go back, now, and calculate the divergence of E directly from Eq. 2.8: f 1 E(r)=- 4JTEo all space 4 / / 2 p (r)dr. !b (2.15) (Originally the integration was over the volume occupied by the charge, but I may as well extend it to all space, since p = 0 in the exterior region anyway.) Noting that the 70 r-dependence is contained in 4 = r - r', we have CHAPTER 2. ELECTROSTATICS V . E = --1 f V· (4)- p(r), , dr . 4JTE"O !z-2 This is precisely the divergence we calculated in Eq. 1.100: Thus V . E = _1_ f 4no 3(r - r')p(r') dr' = ~p(r), (2.16) 4nEO EO which is Gauss's law in differential form (2.14). To recover the integral form (2.13), we run the previous argument in reverse-integrate over a volume and apply the divergence theorem: f V . E dr = 1. E . da = f p dr = Qenc. j EO EO V S V 2.2.3 Applications of Gauss's Law I must interrupt the theoretical development at this point to show you the extraordinary power of Gauss's law, in integral form. When symmetry permits, it affords by far the quickest and easiest way of computing electric fields. I'll illustrate the method with a series of examples

Notice the subtle but crucial role played by path independence (or, equivalently, the fact that V x E = 0) in this argument. Ifthe line inregfal of E depended on the path taken, then the "definition" of V, Eq. 2.21, would be non~ense. It simply would not define a function, since changing the path ~ould alter the value of V (r). By the way, don't let the minus sign in Eq. 2.23 distract you; it carries over from 2.21 and is largely a matter of convention. Problem 2.20 One of these is an impossible electrostatic field. Which one? (a) E = k[xy x+ 2yz5' + 3xd]; (b) E = k[i x+ (2xy + z2) Y+2yz z]. Here k is a constant with the appropriate units. For the possible one, find the potential, using the origin as your reference point. Check your answer by computing VV. [Hint: You must select a specific path to integrate along. It doesn't matter what path you choose, since the answer is path-independent, but you simply cannot integrate unless you have a particular path in mind.] 2.3.2 Comments on Potential (i) The Qflme. The word "potential" is a hideous misnomer because it inevitably reminds you ofpotential energy. This is particularly confusing, because there is a connection between "potential" and "potential energy," as you will see in Sect. 2.4. I'm sorry that it is impossible to escape this word. The best I can do is to insist once and for all that "potential" and "potential energy" are completely different terms and should, by all rights, have different names. Incidentally, a surface over which the potential is constant is called an equipotential. (ii) Advantage of the potential formulation. If you know V, you can easily get E-just take the gradient: E = - VV. This is quite extraordinary when you stop to think about it, for E is a vector quantity (three components), but V is a scalar (one component). How can one function possibly contain all the information that three independent functions carry? The answer is that the three components of E are not really as independent as they look; in fact, they are explicitly interrelated by the very condition we started with, V x E = O. In terms of components, aEy az ' This brings us back to my observation at the beginning of Sect. 2.3.1: E is a very special kind of vector. What the potential formulation does is to exploit this feature to maximum advantage, reducing a vector problem down to a scalar one, in which there is no need to fuss with components.

Setting the reference point at infinity, the potential of a point charge q at the origin is -1 1r q , 1 q I r VCr) = -- -dr = --- 41TEO 00 r,2 41T EO r' 00 q 41TEO r (You see here the special virtue of using infinity for the reference point: it kills the lower limit on the integral.) Notice the sign of V; presumably the conventional minus sign in the definition of V (Eq. 2.21) was chosen precisely in order to make the potential of a positive charge come out positive. It is useful to remember that regions of positive charge are potential "hills," regions of negative charge are potential "valleys," and the electric field points "downhill," from plus toward minus. In general, the potential of a point charge q is 1 q VCr) = --, 41TEO (2.26) where~, as always, is the distance from the charge to r (Fig. 2.32). Invoking the superposition principle, then, the potential of a collection of charges is 1 n q; V(r) = - I:-, 41TEO ;=1 ~; or, for a continuous distribution, VCr) = - 1 f -dq. 1 41TEO In particular, for a volume charge, it's VCr) = _1_ f per') dr'. 41Tfo (2.27) (2.28) (2.29) This is the equation we were looking for, telling us how to compute V when we know p; it is, if you like, the "solution" to Poisson's equation, for a localized charge distribution.5 I

So Gauss's law becomes f (V· E)dr = f (:a) dr. V V And since this holds for any volume, the integrands must be equal: 69 (2.14) Equation 2.14 carries the same message as Eq. 2.13; it is Gauss's law in differential form. The differential version is tidier, but the integral form has the advantage in that it accommodates point, line, and surface charges more naturally.

Such diagrams are also convenient for representing more complicated fields. Of course, the number of lines you draw depends on how energetic you are (and how sharp your pencil is), though you ought to inciude enough to get an accurate sense of the field, and you must be consistent: If charge q gets 8 lines, then 2q deserves 16. And you must space them fairly-they emanate from a point charge symmetrically in all directions. Field lines begin on positive charges and end on negative ones; they cannot simply terminate in midair, though they may extend out to infinity. Moreover, field lines can never cross-at the intersection, the field would have two different ditections at once! With all this in mind, it is easy to sketch the field of any simple configuration of point charges: Begin by drawing the lines in the neighborhood of each charge, and then connect them up or extend them to infinity (Figs. 2.13 and 2.14)

The fundamental problem electromagnetic theory hopes to solve is this (Fig. 2.1): We have some electric charges, q], q2, q3, ... (call them source charges); what force do they exert on another charge, Q (call it the test charge)? The positions ofthe source charges are given (as functions of time); the trajectory of the test particle is to be calculated. In general, both the source charges and the test charge are in motion. The solution to this problem is facilitated by the principle ofsuperposition, which states that the interaction betWeen any two charges is completely unaffected by the presence of others. This means that to determine the force on Q, we can first compute the force F!, due to q! alone (ignoring all the others); then we compute the force F2, due to q2 alone; and so on. Finally, we take the vector sum of all these individual forces: F = F! +F2 +F3 + ... Thus, if we can find the force on Q due to a single source charge q, we are, in principle, done (the rest is just a question of repeating the same operation over and over, and adding it all up).! Well, at first sight this sounds very easy: Why don't I just write down the formula for the force on Q due to q, and be done with it? I could, and in Chapter 10 I shall, but you would be shocked to see it at this stage, for not only does the force on Q depend on the separation distance 1- between the charges (Fig. 2.2), it also depends on both their velocities and on the acceleration of q. Moreover, it is not the position, velocity, and acceleration of q right now that matter: Electromagnetic "news" travels at the speed of light, so what concerns Q is the position, velocity, and acceleration q had at some earlier time, when the message left

Therefore, in spite of the fact that the basic question ("What is the force on Q due to q?") is easy to state, it does not pay to confront it head on; rather, we shall go at it by stages. In the meantime, the theory we develop will permit the solution of more subtle electromagnetic problems that do not present themselves in quite this simple format. To begin with, we shall consider the special case of electrostatics in which all the source charges are stationary (though the test charge may be moving). 2.1.2 Coulomb's Law What is the force on a test charge Q due to a single point charge q which is at rest a distance 1- away? The answer (based on experiments) is given by Coulomb's law: 1 qQA F= ----4. 47TEO 1-2 (2.1) The constant EO is called the permitivity of free space. In SI units, where force is in Newtons (N), distance in meters (m), and charge in coulombs (C), -12 C2 EO = 8.85 x 10 --2' N·m In words, the force is proportional to the product of the charges and inversely proportional to the square of the separation distance. As always (Sect. 1.1.4),4 is the separation vector from r/ (the location of q) to r (the location of Q): 4 = r - r/; (2.2) 1- is its magnitude, and 4 is its direction. The force points along the line from q to Q; it is repulsive if q and Q have the same sign, and attractive if their signs are opposite. Coulomb's law and the principle of superposition constitute the physical input for electrostatics-the rest, except for some special properties of matter, is mathematical elaboration of these fundamental rules.

Work and Energy in Electrostatics 2.4.1 The Work Done to Move a Charge Suppose you have a stationary configuration ofsource 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 = QE; the force you must exert, in opposition to this electrical force, is - QE. (If the sign bothers you, think about lifting a brick: Gravity exerts a force mg downward, but you exert a force mg upward. Of course, you could apply an even greater force-then the brick would accelerate, and part

You may have noticed, in studying Exs. 2.4 and 2.5, or working problems such as 2.7, 2.11, and 2.16, that the electric field always undergoes a discontinUity when you cross a surface charge a. In fact, it is a simple matter to find the amount by which E changes at such a boundary. Suppose we draw a wafer-thin Gaussian pillbox, extending just barely over the edge in each direction (Fig. 2.36). Gauss's law states that I E. da = ~Qenc = ~aA, EO EO S where A is the area of the pillbox lid. (If a varies from point to point or the surface is curved, we must pick A to be extremely small.) Now, the sides of the pillbox contribute nothing to the flux, in the limit as the thickness E goes to zero, so we are left with 1.. 1.. 1 Eabove - Ebelow = -a, EO (2.31) where Ei1bove denotes the component of E that is perpendicular to the surface immediately above, and Etelow is the same, only just below the surface. For consistency, we let "upward" be the positive direction for both. Conclusion: The normal component of E is discontinuous by an amount a / EO at any boundary. In particular, where there is no surface charge, E1.. is continuous, as for instance at the surface of a uniformly charged solid sphere. The tangential component of E, by contrast, is always continuous. For if we apply Eq.2.19, f E· dl = 0, to the thin rectangular loop of Fig. 2.37, the ends give nothing (as E ---+ 0), and the sides . ( II II give Eabovel - Ebelowl), so E" -E" above - below

and the magnitude of E is constant over the Gaussian surface, so it comes outside the integral: JlEI da = lEIJda = IEI4JTr 2 . S S Thus 2 1 IEI4JTr = -q, EO or 1 q A E= ---r. 4JTEO r 2 Notice a remarkable feature of this result: The field outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center.

from point to point and is determined by the configuration of source charges; physically, E(r) is the force per unit charge that would be exerted on a test charge, if you were to place one at P. What exactly is an electric field? I have deliberately begun with what you might call the "minimal" interpretation of E, as an intermediate step in the calculation of electric forces. But I encourage you to think of the field as a "real" physical entity, filling the space in the neighborhood of any electric charge. Maxwell himself came to believe that electric and magnetic fields represented actual stresses and strains in an invisible primordial jellylike "ether." Special relativity has forced us to abandon the notion of ether, and with it Maxwell's mechanical interpretation of electromagnetic fields. (It is even possible, though cumbersome, to formulate classical electrodynamics as an "action-at-a-distance" theory, and dispense with the field concept altogether.) I can't tell you, then, what a field is-only how to calculate it and what it can do for you once you've got it

invite you to compare Eq. 2.29 with the corresponding formula for the electric field in terms of p (Eq. 2.8): I !p(r')A , E(r) = -- --2-Itdr. 4JrEo !b The main point to notice is that the pesky unit vector i is now missing, so there is no need to worry about components. Incidentally, the potentials of line and surface charges are _1_ ! ).(r') dl' and 4JrEo !b _1_! a(r') da'. 4JrEo !b (2.30) I should warn you that everything in this section is predicated on the assumption that the reference point is at infinity. This is hardly apparent in Eq. 2.29, but remember that we got that equation from the potential of a point charge at the origin, (1/4JrEO) (q / r), which is valid only when 0 = 00. If you try to apply these formulas to one of those artificial problems in which the charge itself extends to infinity, the integral will diverge.

of your effort would be "wasted" generating kinetic energy. What we're interested in here is the minimum force you must exert to do the job.) The work is therefore w = {b F. dl = _Q (b E. dl = Q[V(b) - Yea)]. Ja Ja Notice that the answer is independent of the path you take from a to b; in mechanics, then, we would call the electrostatic force "conservative." Dividing through by Q, we have W V(b) - Yea) = -. Q (2.38) In words, the potential difference between points a and b is equal to the work per unit charge required to carry a particle from a to b. In particular, if you want to bring the charge Q in from far away and stick it at point r, the work you must do is W = Q[V(r) - V(oo)], so, if you have set the reference point at infinity, W = QV(r). (2.39) In this sense potential is potential energy (the work it takes to create the system) per unit charge Gust as the field is the force per unit charge). 2.4.2 The Energy of a Point Charge Distribution 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, ql, takes no work, since there is no field yet to fight against. Now bring in qz. According to Eq. 2.39, this will cost you qZ VI (rz), where VI is the potential due to ql, and rz is the place we're putting qz:

so if you double p, you double E. (Wait a minute! How do we know that doubling Q (and also -Q) simply doubles p? Maybe the charge moves around into a completely different configuration, quadrupling p in some places and halving it in others, just so the total charge on each conductor is doubled. The fact is that this concern is unwarranted-doubling Q does double p everywhere; it doesn't shift the charge around. The proof of this will come in Chapter 3; for now you'll just have to believe me.) Since E is proportional to Q, so also is Y. The constant of proportionality is called the capacitance of the arrangement: . C= Q - y' Capacitance is a purely geometrical quantity, determined by the sizes, shapes, and separation ofthe two conductors. In SI units, C is measured in farads (F); a farad is a coulomb-per-volt. Actually, this turns out to be inconveniently large;9 more practical units are the microfarad (10-6 F) and the picofarad (10- 12 F). Notice that Y is, by definition, the potential of the positive conductor less that of the negative one; likewise, Q is the charge of the positive conductor. Accordingly, capacitance is an intrinsically positive quantity. (By the way, you will occasionally hear someone speak of the capacitance of a single conductor. In this case the "second conductor," with the negative charge, is an imaginary spherical shell of infinite radius surrounding the one conductor. It contributes nothing to the field, so the capacitance is given by Eq. 2.53, where Y is the potential with infinity as the reference point.)

whereas the sides contribute nothing. Thus I 2A lEI = -e>A, EO or E=~1i 2Eo (2.17) where Ii is a unit vector pointing away from the surface. In Prob. 2.6, you obtained this same result by a much more labori0!1s method. It seems surprising, at first, that the field of an infinite plane is independent ofhowfar away you are. What about the 1/r2 ill Coulomb's law? Well, the point is that as you move farther and farther away from the plane, more and more charge comes into your "field of view" (a cone shape extending out from your eye), and this compensates for the diminishing influence of any particular piece. The electric field of a sphere falls off like 1/r 2; the electric field of an infinite line falls off like 1/r; and the electric field of an infinite plane does not fall off at all.

vwhereas the sides contribute nothing. Thus I 2A lEI = -e>A, EO or E=~1i 2Eo (2.17) where Ii is a unit vector pointing away from the surface. In Prob. 2.6, you obtained this same result by a much more labori0!1s method. It seems surprising, at first, that the field of an infinite plane is independent ofhowfar away you are. What about the 1/r2 ill Coulomb's law? Well, the point is that as you move farther and farther away from the plane, more and more charge comes into your "field of view" (a cone shape extending out from your eye), and this compensates for the diminishing influence of any particular piece. The electric field of a sphere falls off like 1/r 2; the electric field of an infinite line falls off like 1/r; and the electric field of an infinite plane does not fall off at all.

where Ell stands for the components of E parallel to the surface. The boundary conditions on E (Eqs. 2.31 and 2.32) can be combined into a single formula: a A Eabove - Ebelow = - n, EO (2.33) where nis a unit vector perpendicular to the surface, pointing from "below" to "above.,,6 The potential, meanwhile, is continuous across any boundary (Fig. 2.38), since Vabove - Vbelow = - l b E . dl; as the path length shrinks to zero, so too does the integral: Vabove = Vbelow

which puts out a field (0-/2Eo) on either side, pointing away from the surface (Fig. 2.50). Thus, 0- A E above = E other + -2n, EO 0- A Ebelow = Eother - -n, 2Eo and hence 1 E other = "2 (E above + E below) = E average· Averaging is really just a device for removing the contribution of the patch itself. That argument applies to any surface charge; in the particular case of a conductor, the field is zero inside and (0-/EO)i'i outside (Eq. 2.48), so the average is (0-/2Eo)i'i, and the force per unit area is 1 2 A f = -0- n. (2.51) 2Eo This amounts to an outward electrostatic pressure on the surface, tending to draw the conductor into the field, regardless of the sign of 0-. Expressing the pressure in terms of the field just outside the surface,

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