Electric Fields Around Conductors Part 1, Solution Manual

Electric fields around conductors Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin. + (a) + In or out - + The charge distributions are shown in Fig. 64. In both cases, there is negative charge on the inside surface of the inner “ring” (due to the attraction to the charge q) and positive charge on the outside surface of the outer ring (due to the self repulsion of the leftover positive charge). In the first case, the charge q is outside the conducting shell, so there must be zero field inside. Consistent with this, there is no charge on the inner side of the surface that touches the hollow interior of the conductor. (If there were such a charge, we could draw a Gaussian surface that lies partially inside the metal of the conductor, and partially in the interior of the conducting shell, to show that there would be a nonzero field in the interior.) In the second case, the charge q is inside the conducting shell, so there is a nonzero field in the interior. Consistent with this, there is nonzero charge (the negative charge shown) on the inner side of the surface that touches the hollow interior of the conductor. The positive charge is still outside, as it was in the first case. The first case is consistent with the fact that there is no electric field in the hollow interior of a conducting shell containing no charge, while the second case doesn’t apply because there is charge inside. Gravity screen The electric field can be “blocked” because of the existence of charges of opposite signs. As seen in Fig. 3.8, these oppositely-signed charges set up a compensating electric field. If only positive charges existed, conductors couldn’t block electric fields. For this reason, since only one sign of gravitational mass exists, it is impossible to block the gravitational field. If negative-mass particles existed, then in theory it would be possible to block a gravitational field. For example, if we have a point mass located outside a spherical shell, we could put some negative mass on the near size of the shell. This mass would + - - + - - + + -- + (b) + + + + - + + + -- - - - + Figure 64 + + repel a positive-mass particle in the interior, canceling the attraction from the original positive point mass. However, there is one aspect in which the gravitational case differs fundamentally from the electrical case. In the latter, like charges repel; whereas in the former, like masses attract. This means that in the gravitational case, like masses would attract each other and collapse down to a point, if they were allowed to move freely. So we would have to bolt them down in the desired distribution. Two concentric shells (a) The charge distributions and field lines are shown (roughly) in Fig. 65. Let the four surfaces be labeled 1, 2, 3, 4, starting from the innermost one. There is charge −q on surface 1. This is true because the field is zero inside the metal of the conductor, so a spherical Gaussian surface drawn inside the metal of the inner conductor has no flux, so the net charge enclosed in the sphere must be zero. The negative surface charge density on surface 1 is higher near the off-center point charge. Since the inner conductor is neutral, a charge +q must reside on surface 2. This surface charge density is spherically symmetric, because it feels no field from the charges inside (or outside) of it, due to the zero field inside the metal of the conductors. a) +E= 0 + +- 3 - 2 + + +-1 - + + -+ -++- + - q q + +- - + + ++ ++ E= 0 4 b) + + + ++ + 4+ + 3 + + - q + E= 0 - - 1 - + + 2 wire - - q + E= 0 Figure 65 By the same Gauss’s-law reasoning, there must be a charge −q on surface 3, because there is zero field inside the metal of the outer conductor. This surface charge density is spherically symmetric. A charge +q is left for surface 4. This surface charge density is actually negative near the outer charge q if that charge is located close enough to the shells; see Exercise 3.49. But in any case the total charge on surface 4 is +q. Between the shells, the field is spherically symmetric, consistent with the spherically symmetric charge densities on surfaces 2 and 3. (b) The shells are now at the same potential, so the field between them must be zero. Therefore, the only difference from the scenario in part (a) is that we just need to erase the field between the shells and erase the charges on surfaces 2 and 3. The charges on surfaces 1 and 4 aren’t affected by this change, because the surfaces 2 and 3 together produced zero field everywhere except between them. Basically, when we connect the shells, the charges on surfaces 2 and 3 simply neutralize each other. Equipotentials Assume that the point charge is positive (the general result is the same if it is negative). Then the near part of the sphere ends up negatively charged, and the far part ends up positively charged. (The sizes of these two regions depend on the distance from the point charge to the sphere.) By continuity, there must therefore be a circle on the sphere where the charge density is zero. But the electric field near the sphere (which is perpendicular to the conducting surface) is given by σ/ϵ0 . So if σ = 0 on the circle, then E = 0 also. The general shapes of the equipotentials are shown in Fig. 66. (The various curves have been chosen to indicate the general features; their potential values aren’t equally spaced.) In this specific case, the distance from the point charge to the sphere has been chosen to be twice the radius of the sphere. The transition from small circles around the point charge to large circles around the whole system takes place via the equipotential curve that heads straight into the sphere and then splits in two, encircling the sphere. At the point where the curve splits, it changes direction abruptly. Since the electric field must be perpendicular to the equipotential surface at every point, this means that the electric field must point in two different directions at the splitting point. The only vector that is perpendicular to two different directions is the zero vector. So this is a second way of seeing why there must be points on the surface of the sphere where the electric field is zero. 3 2 1 conducting sphere 0 -1 -2 -3 -2 -1 0 1 2 3 4 Figure 66 Surface density at a corner Consider first the electric field due to an infinite strip of charge, with width b, infinite length, and negligible thickness. The surface charge density takes on the constant value σ. Let us find the electric field at a point P located a distance x from one edge of the strip, in the plane of the strip. We can slice the strip into narrow rods, and then integrate over the rods. Consider a rod with width dr at a distance r from a given point P ; see Fig. 67. The electric field at P due to the rod is λ/2πϵ0 r, where λ = σ dr. The distance r runs from x to x + b, dr b r P x σ Figure 67 so the total field at P due to the strip is ) ( ∫ x+b σ dr x+b σ . ln E(x) = = x 2πϵ r 2πϵ 0 0 x (227) If x → 0 this result diverges (slowly, like ln x). This divergence is what causes the electric field at the corner of the square tube to diverge, for the following reason. x x P Figure 68 If we treat the corner like an exact point, then a cross section is shown in Fig. 68. The given point P is near the edges of two different strips (the two adjacent faces of the tube). P doesn’t lie exactly in the plane of each strip, but this doesn’t matter. The field from each strip differs from the field in part (a) by a finite additive amount, so it still diverges as x → 0. This is true because if we ignore the “rods” in the strip that are within a distance of, say, 5x from P , then P can be treated as essentially lying in the plane of the remaining part of the strip. The effective value of x is now 6x, but the factor of 6 doesn’t matter; the field still diverges as x → 0. (This reasoning holds for any location near the corner; P need not lie on the line of the angle bisector.) We are concerned only with the component of √ the field that lies along the angle bisector, so this brings in a factor of cos 45◦ = 1/ 2 in the field from each strip. But this doesn’t change the fact that the total field diverges. If we treat the corner more realistically as curved (like a quarter circle), then the above reasoning still applies. Ignoring the nearby part of the charge distribution still leaves us with two strips that each produce an infinite field, in the limit where the radius of curvature of the quarter circle goes to zero (assuming that P is close to the quarter circle, on the order of the radius). If the radius does not go to zero, then the field certainly doesn’t diverge. So the “corner” of the tube needs to be sharp in order for the field to diverge. We have been treating the charge density σ as constant. But in a conducting tube, the density increases near the corners, because of the self-repulsion of the charges. This has the effect of making the field even larger than the above reasoning would imply, so the above conclusion of a diverging field is still valid. Since the conclusion is true for both conducting and nonconducting tubes, the word “conducting” in the statement of the problem could have been omitted. P Figure 69 In the case of a curved corner, if P is very close to the quarter circle (or whatever curve), then we can draw a tiny Gaussian pillbox Fig. 69 to say that the field at P equals σ/ϵ0 . Since we just showed that the field diverges, this implies that the density also diverges at the corner. Intuitively, if it didn’t diverge, then there wouldn’t exist a sufficient force to keep the charges in the straight parts of Fig. 69 from flowing onto the curved part. So this would eventually lead to a very large density at the corner anyway. All of the above reasoning still holds if the cross section of the tube is something other than a square. At any point where the direction of the surface changes abruptly, the field diverges. Even for a polygon with 100 sides, in which the surface bends by only a few degrees at each “corner,” the field still diverges, because when taking the component along the angle bisector, the nonzero trig factor doesn’t change the fact that the total field diverges. If we kick things down a dimension and look at a kink in a wire, the field still diverges (even more quickly). This is true because the field near the end of a uniform stick diverges; Eq. (227) is replaced by ) ( ∫ x+b λ dr 1 1 λ . (228) E(x) = − = 2 x 4πϵ + b r x 4πϵ 0 0 x This diverges as x → 0. In the case of a cone, a slightly different calculation is required, but the field still diverges. See Problem 1.3. Zero flow The point is that although the field is weaker near the larger sphere, it has an appreciable size over a larger distance than does the field from the smaller sphere. The fields at the surfaces are proportional to Q/R2 ∝ R/R2 = 1/R. And the fields fall off on a distance scale of R, because at a radius of 2R, the field has decreased by a factor 1/22 , etc. These two effects cancel. We can be quantitative. As in Problem 3.10, the thin wire has essentially zero capacitance, so charge can’t pile up on it. But a tiny bit can, so as mentioned in the solution to Problem 3.10, we effectively have a rigid stick of (a tiny bit of) charge extending from one sphere to the other. Assuming constant density λ, the repulsive force on ∫ D−R the stick from the smaller sphere is R1 2 E1 λ dr, where D is the distance between the centers of the spheres, and where E1 (r) is the field due to the smaller sphere. ∫ D−R Likewise, the repulsive force from the larger sphere is R2 1 E2 λ dr. If the spheres are far apart, we can replace the upper limits of these integrations by ∞; the fields become negligible at large distances. If we set these two forces equal to each other and cancel the λ’s, we simply have the statement that the potentials of each shell, relative to infinity, are equal. In a sense, we can think of many weak people forcing the charge away from the larger sphere, with only a few strong people forcing it away from the small sphere. The two total forces exactly cancel, and no charge moves. y= 0 y =s A charge between two plates Consider the Gaussian surface shown in Fig. 70, which has very large extent in the x and z directions. Two of its faces lie inside the metal of the two plates. The field is zero inside the metal of the plates; and between the plates the field is negligible at points far away from the charge. So there is zero flux through the Gaussian surface. By Gauss’s law, the total charge inside must be zero, which implies that the total charge on the inner surfaces of the plates must be −Q. Dividing the charge Q into many smaller charges, all on the plane y = s, yields the same total charge on each plate, by superposition. We can take the continuum limit and smear out the charge Q uniformly onto a sheet with a large area A at y = b. The surface charge density on this sheet is σ = Q/A (this is the sum of the densities on its two surfaces). If we can find the total charges on the two plates in this scenario, then we will have found the total charges on the two plates in the original scenario involving the point charge Q. Call the charges on (the inner surface of) each plate Q1 and Q2 . The densities are then σ1 = Q1 /A and σ2 = Q2 /A, so Q1 /Q2 = σ1 /σ2 . The same area A applies to the two plates because we are assuming A is large. The edge effects will be negligible, so we can assume that σ1 and σ2 are essentially uniform over the area A. (And σ is uniform, by construction.) From the standard argument using Gauss’s law with one face of the Gaussian surface lying inside the metal of a conductor, the fields in the two different regions are E1 = σ1 /ϵ0 and E2 = σ2 /ϵ0 . So E1 /E2 = σ1 /σ2 . But the two plates are at the same potential, so we also know that E1 b = E2 (s − b), because these are the differences in potential from the middle sheet. Hence E1 /E2 = (s − b)/b. So we have Q 1 σ1 E1 s−b = = = b Q2 σ2 E2 =⇒ Q1 s−b . = b Q2 (229) Q b left plate right plate Figure 70 But we also have Q1 + Q2 = −Q. Solving these two equations for Q1 and Q2 gives Q1 = −Q s−b s and b Q2 = −Q . s (230) If b ≪ s, then we have Q1 ≈ −Q and Q2 ≈ 0, as expected. In general, the charges are in the inverse ratio of the distances from the plates to the intermediate sheet (or to the given point charge Q). Two charges and a plane Q l -Q y Q First note that such a location must exist, due to a continuity argument: If the −Q charge is placed only slightly below the fixed Q charge, the upward attractive force from the Q charge will dominate. But if the −Q charge is placed only slightly above the conducting plane, the downward attractive force from the +Q image charge will dominate. So somewhere in between, the force on the −Q charge must be zero. Let y be the distance from the −Q charge to the plane. The field above the plane due to the two given charges along with the induced charge on the plane is identical to the field due to the two given charges along with the two image charges below the plane shown in Fig. 71. The given −Q charge feels the fields due to the other three charges. Taking upward to be positive, the force on the given −Q charge is ) ( 1 Q2 1 1 . (231) F = − + (ℓ + y )2 4πϵ0 (ℓ − y)2 (2y)2 Setting this equal to zero yields -Q Figure 71 1 2(ℓ2 + y 2 ) = 2 (ℓ − y 2 )2 4y 2 =⇒ 7y 4 + 10ℓ2 y 2 − ℓ4 = 0. (232) This is a quadratic equation in y 2 . We are concerned with the positive root (since y 2 is positive), which is √ (−5 + 4 2)ℓ2 ≈ (0.0938)ℓ2 =⇒ y = (0.306)ℓ. (233) y2 = 7 A wire above the earth Let L = 200 m, h = 5 m, and λ = 10−5 C/m. By superposition, the relevant image charge is an oppositely-charged wire of the same length below the surface of the earth. Because L is much larger than h, we can consider (except near the ends of the wire) the wires to be of infinite length, as far as finding the field goes. The field at the surface of the earth is due to both of the wires, so it points downward with magnitude Esurface = 2 · V 10−5 C/m λ ) . = 7.2 · 104 = ( s2 C2 m 2πϵ0 h π 8.85 · 10−12 kg m3 (5 m) (234) The electrical force on the given wire is the force due to the field arising from the imagecharge wire. The total charge on the given wire is q = λL = (10−5 C/m)(200 m) = 2 · 10−3 C. Over nearly the whole length of the wire, the field due to the imagecharge wire is essentially λ/2πϵ0 (2h) = 1.8 · 104 V/m, directed downward. This is a quarter of the field we found above, which involved two wires and half the distance. Neglecting the decrease in field near the ends of the wire, the force on the given wire is qE = (2 · 10−3 C)(1.8 · 104 V/m) = 36 N, directed downward. In terms of the various parameters, this force is qE = (λL)(λ/2πϵ0 (2h)) = (λ2 /4πϵ0 )(L/h). Remark: The above result is a good approximation in the limit where L ≫ h. Let’s try to get a handle on the error involved. Since we assumed that the field was uniform over the length of the wire, what we actually calculated was the force on a finite wire due to an infinite wire (Fig. 72(a)). This is larger than the force between two finite wires (Fig. 72(b)) because of the two extra half-infinite wires (Fig. 72(c)) that need to be added to the lower finite wire to make the infinite wire. (a) (b) (c) (d) Figure 72 As as exercise, you can show that the vertical component of the force between the two halfinfinite wires in Fig. 72(d) is λ2 /4πϵ0 , which is independent of the separation (this follows from a dimensional-analysis argument). To a good approximation, the original finite wire can be treated as half-infinite for this purpose. (The error is of order (λ2 /4πϵ0 )(h/L).) So the force between the finite wire and the two half-infinite wires in Fig. 72(c) equals 2λ2 /4πϵ0 . This is the correction we need to make to our original result. So the actual force between the two finite wires in Fig. 72(b) is ( ) λ2 L λ2 L 2λ2 2h = − 1− , (235) 4πϵ0 h 4πϵ0 4πϵ0 h L up to corrections of higher order in h/L. In the present setup, 2h/L equals 1/20. So by ignoring the end effects, we over-estimated the force by about 5%. Direction of the force There are two negative image charges on the other side of the plane, at the mirrorimage locations. For very large z values of the top charge q, the lower q and its image charge −q look like a dipole from afar, which has a repulsive (upward) field that falls off like 1/z 3 . But the attractive (downward) field from the other image charge −q behaves like 1/(2z)2 . This has a smaller power of z in the denominator, so it dominates for large z. The force on the top charge q is therefore downward for large z. So the answer to the stated question is “No.” Q Horizontal field line There are many different Gaussian surfaces that can be used to solve this problem. One is shown in Fig.73. The bottom of the surface can be chosen to lie either inside the material of the conducting plane, or below it; in either case the field is zero there. The rest of the surface follows the field lines that start out horizontal. The exception is right near the point charge, where the surface takes the form of a tiny hemisphere. There is no flux through the surface except through the hemisphere, because everywhere else the field is either zero or parallel to the surface. Figure 73 The flux into the surface is (Q/2)/ϵ0 because exactly half of the Q/ϵ0 flux that passes outward through a tiny sphere around the charge Q (in a spherically symmetric manner very close to the charge) passes though the bottom half of the sphere. Since this flux passes into our Gaussian surface, Gauss’s law tells us that there must be a charge of −Q/2 inside. The only place this charge can be is on the surface of the plane. So our task is to find the radius R of the circle that contains half of the total charge −Q on the plane. ∫R Using the density σ given in Eq. (3.4), the requirement −Q/2 = 0 σ2πr dr becomes 1 = 2 Therefore, ∫ 0 R R  −h hr dr  =1−√ h = √ . (r2 + h2 )3/2 r2 + h2 0 R 2 + h2 √ h R 2 + h2 = 1 2 =⇒ R = √ 3 h. (236) (237) Remark: The above reasoning works for any starting angle of the field lines, not just horizontal. If we measure θ with respect to the upward vertical, then as an exercise you can shown that a field line that starts out at an angle θ meets the plane at a radius R = √ h 3 + 2 cos θ − cos2 θ/(1 − cos θ). (As a sub-problem, you will need to show that the fraction of the total surface area of a sphere that lies in a spherical cap subtended by the cone with therefore subtends a fraction (1 + cos θ)/2.) For half-angle θ is (1 − cos θ)/2. The remainder √ θ = π/2, this correctly gives R = 3 h. And for θ = 0 and π it gives, respectively, R = ∞ and 0, as it should. -q r1 r1 r2 r2 q Interestingly, for θ close to π (call it π − ϵ), that is, for field lines that start off pointing √ nearly straight downward, you can show that R ≈ hϵ/ 2. This is correctly smaller than hϵ, because that would be where the field line would hit the plane if the line were exactly straight, whereas we know that it must bend so that it ends up vertical when it meets the √ plane. But the 1/ 2 factor isn’t obvious. Point charge near a corner q -q Figure 74 The two equipotential surfaces are the planes shown in Fig. ∑ 74. The potential is zero at every point on these planes, because at any such point, (±q/r) = 0 by symmetry. Equivalently, the electric field is perpendicular to the planes at every point. This can be seen by grouping the four charges into two dipoles (with charges on opposite sides of a given plane); the field from each dipole is perpendicular to the plane at every point on the plane. By superposition, the field from both dipoles is also perpendicular to the plane at every point on the plane. 73 This setup satisfies the boundary conditions of the setup consisting of a point charge and a right-angled sheet. The field is shown (roughly) in Fig. 75. This field is similar to two copies of Fig. 3.10(b), tilted at right angles with respect to each other, but with the lines pushed away from the corner. The field at the center of the square of the original four charges is zero, so the field at the corner of the metal sheet is likewise zero. This implies that the surface density is zero at the corner. In order for this method to work, we need to divide space into an even number of identical wedges (call it 2N ), in order to have an alternating ring of charges. So it won’t work unless the angle θ at the bend in the sheet is of the form θ = 2π/2N = π/N , where N is an integer. It won’t work, for example, with θ = 120◦ = 2π/3. -q q q -q Figure 75 In the setup with the right-angled corner, note that the total charge on the two parts of the sheet must be −q (assuming the given charge is q). So the charge on each part must be −q/2. It is possible to verify this explicitly by writing down the field E (and hence density σ = ϵ0 E) as a function of the two coordinates associated with the sheet, and then performing (ideally with a computer) the double integral of σ over the half-infinite sheet. If the given charge q is located a distance d from each sheet of the right-angled corner, then by looking at the forces from the three√image charges, you can show that the force on the given charge q is (q 2 /32πϵ0 d2 )(2 2 − 1), directed toward the corner. 2d z Q Images from three planes The required image charges are at the other seven corners of a cube of side 2d, as shown in Fig. 76. This configuration satisfies the condition of equipotential surfaces where the planes∑ are; the potential is zero at every point on these planes, because at any such point, (±Q/r) = 0 by symmetry. Equivalently, the total electric field is perpendicular to all of the planes at every point. This can be seen by grouping the eight charges into four dipoles (with charges on opposite sides of a given plane); the field from each dipole is perpendicular to the plane at every point on the plane. By superposition, the field from all four dipoles is also perpendicular to the plane at every point on the plane. From the symmetry of the setup, the net force on Q is directed toward (or away) from the origin (the center of the cube of side 2d). So we need compute only the force components in that direction. There are three classes of charges: √ • Three charges −Q at a distance 2d make an angle cos−1 (1/ 3) with the direction toward √ the origin. This follows from the fact that the diagonal of the cube has length 3(2d). Alternatively, you can find the cosine by using the two standard expressions for the dot product. We therefore have (ignoring the 4πϵ0 ), √ Q2 1 3 Q2 F1 = 3 ·√ = (toward origin). (238) 2 4 d2 (2d) 3 √ √ √ • Three charges Q at a distance 2 2 d make an angle cos−1 ( 2/ 3) with the direction toward the origin. So √ √ Q2 2 3 Q2 F2 = 3 √ (away from origin). (239) · √ = √ 2 (2 2d)2 3 4 2 d √ • One charge −Q at a distance 2 3 d is located in line with the origin. So F3 = Q2 Q2 √ = . 12d2 (2 3 d)2 (toward origin) (240) x Figure 76 y The total force on Q is therefore (bringing back in the 4πϵ0 ) ) (√ √ (0.210) Q2 3 1 Q2 3 1 ≈ (toward origin). − √ + F = 2 d 4 12 4πϵ0 4πϵ0 d2 4 2 (241) Force on a charge between two planes If the charge is very close to the right plane, as in Fig. 12.53, we see that the force on the charge comes from the image charge nearby on its right, plus an infinite number of dipoles. Two these dipoles are at distances of approximately 2ℓ, two are at 4ℓ, and so on. The strength of the dipoles is p = q(2b). From Eq. (2.36), the field from a dipole, along the axis, is E = 2p/4πϵ0 r3 . These fields all point to the left. The total force on the given charge is therefore approximately equal to ( ) ( ) 2 q(2b) q2 1 1 1 −2·q + + F = + · · · 2 3 3 4πϵ0 (2b) (2ℓ) (4ℓ) (6ℓ)3 4πϵ0 ) ( 1 q2 b q2 1 (242) 1 + 3 + 3 + ··· . − = 3 4πϵ0 ℓ3 16πϵ0 b2 2 The factor is parenthesis is approximately equal to 1.2. Note that the total force from the dipoles is smaller than the force from the closest image charge by a factor of order (b/ℓ)3 . Two of these powers of b/ℓ come from the fact that the distances from the given charge to the dipoles are on the order of ℓ instead of b. And the third power comes from the dipole effect of taking the difference between nearly-canceling forces. You can also calculate the total force by looking at the forces from the positive and negative image charges separately. From Fig. 12.53, the forces from the positive charges cancel, because they are symmetrically located with respect to the given charge. The force from the closest negative charge is q 2 /4πϵ0 (2b)2 , directed to the right. The forces from the other negative charges nearly cancel in pairs. The sum of the forces from all these pairs points to the left, and its magnitude can be written as ) ∞ ( q2 ∑ 1 1 − Fneg = 2 4πϵ0 n=1 (2nℓ − 2b)2 (2nℓ + 2b) ) ( ∞ ∑ 1 1 q2 1 = − 2 2 2 2 16πϵ0 ℓ (1 + b/nℓ) n (1 − b/nℓ) n=1 (( ) ( )) ∞ ∑ 2b 2b q2 1 1+ − 1− ≈ 2 2 n nℓ nℓ 16πϵ0 ℓ n=1 = ∞ q2 b ∑ 1 , 3 4πϵ0 ℓ n=1 n3 (243) in agreement with the dipole term in Eq. (242). Charge on each plane (a) At a given point P on the right plane in Fig. 12.53, we need to add up the x components of the fields due to the real charge and all the image charges. The two nearest charges on either √ side of the right plane (the real charge and image charge 1) are a distance b2 + r2 away from P . So the magnitude of the field from each charge is q/4πϵ0 (b2 + r2 ). Taking the x component brings in a factor √ of b/ b2 + r2 . Both charges produce a positive x component, so that brings in a factor of 2. Putting it all together gives the first term in Eq. (3.41). Now consider images charges 2 and 4 in Fig. 12.53. They both are a distance 2ℓ − b from the right plane, so we simply need to replace b with 2ℓ − b in the above reasoning. And we also need to add on a minus sign since the x component is negative. So we obtain the first term in the sum in Eq. (3.41) with n = 1. Similarly, charges 3 and 5 both are a distance 2ℓ + b from the right plane, so they yield the second term in the sum with n = 1. In the same manner, charges 6 and 8 with distances 4ℓ − b, and charges 7 and 9 with distances 4ℓ + b, yield the n = 2 terms. And so on. (b) As in Eq. (3.5), the integral of the first term in Eq. (3.41) (after dividing by −4π to obtain the density) is − 1 4π ∫ R 0 (b2 R  2qb qb  = −q, 2πr dr = 2  2 1/2 2 3/2 (b + r ) +r ) 0 (244) where we have set R = ∞, which causes no problem with this term. In the same manner, the integral arising from the pair of terms with a particular value of n is ) R (  q(2nℓ − b) q(2nℓ + b)  (245) − ( )1/2  . )1/2 + ( 2 2 2 2 (2nℓ + b) + r (2nℓ − b) + r 0 As stated in the problem, if we set R = ∞ these two terms equal ±q, so they cancel. We’ll therefore let R be a large but finite distance. Then when the first term is evaluated at R it can be approximated as (dropping the b2 term) ( ) q(2nℓ − b) q(2nℓ − b) 1 4nℓb −( , (246) )1/2 1 + 2 2 2 )1/2 ≈ − ( 2 4n ℓ + R 2 2 2 2 2 2 4n ℓ + R (4n ℓ + R ) − 4nℓb where we have used 1/(1 − ϵ)1/2 ≈ 1 + ϵ/2. The terms that don’t involve b (or that involve b2 ) will cancel with the corresponding terms in the second term in Eq. (245), which looks the same except for the overall minus sign and the replacement b → −b. So we care only about the b terms. You can show that their sum for a given value of n is 2qbR2 /(4n2 ℓ2 + R2 )3/2 . The factor of 2 out front comes from the fact that there are two terms in Eq. (245). We must now sum this over n. For large R, the terms change slowly with n, so we can approximate the sum by an integral. Let’s relabel n as z. In the original sum over n, we can multiply each term by dn, because dn simply equals 1 since n runs over the integers. We can then replace dn by dz. Integrating over z from 0 (although the exact starting point doesn’t matter) to ∞ gives (you should verify this integral) ∞ ∫ ∞  2qbR2 dz 2qbz  = qb . (247) =√ 2 ℓ2 + R2 )3/2 2 2 2 ℓ (4z 4ℓ z + R  0 0 Remembering to include the −q charge in Eq. (244), the total charge on the right plane is −q + qb/ℓ = −q(ℓ − b)/ℓ. This equals −q if b = 0, and 0 if b = ℓ. These makes sense, because the left plane or right plane, respectively, is irrelevant in these two cases. The charge on the left plane (is obtained)by letting b → ℓ−b throughout the above calculation, which yields −q ℓ − (ℓ − b) /ℓ = −qb/ℓ. Alternatively, we know that the total charge on the inner surfaces of the two planes must be −q (by using a Gaussian surface with two faces lying inside the metal of the conducting planes, where the field is zero). So −q + qb/ℓ on the right plane implies −qb/ℓ on the left plane. Sphere and plane image charges Let the shell have radius R. Consider a point charge Q located at x = A, where A = R + h with h ≪ R. Then from Problem 3.13, the image charge −q inside the shell equals −QR/A = −QR/(R + h) ≈ −Q, and its location is x= R2 R R2 = ≈ R(1 − h/R) = R − h. = R + h 1 + h/R A (248) Very close to a spherical shell, the shell looks locally like a plane. So what we essentially have here is two charges ±Q located on either side of a plane, a distance h from it. This is exactly what our image-charge setup looked like in the case of a plane. The same type of reasoning holds if the given charge lies inside the shell. The real and image charges have now simply switched sides of the plane. Bump on a plane Q R/2 -Q/2 Q/2 -Q Figure 77 The point charge Q is a distance A = 2R from the center of the hemisphere. So Problem 3.13 tells us that if an image charge of −q = −QR/A = −Q/2 is located a distance a = R2 /A = R/2 above the center of the hemisphere, the whole sphere will be at zero potential (even though we don’t care about the bottom half). But we still need to make the whole plane an equipotential. We can do this by adding the opposites of the two existing charges (one real and one image), but now below the plane, as shown in Fig. 77. The two new image charges still have the sphere as an equipotential surface. And all four charges have the whole plane as an equipotential (at zero potential). This is clear if we group the charges into two pairs: the ones at y = ±R/2, and the ones at y = ±2R. We have actually created an equipotential surface consisting of the union of the plane and the sphere, but the bottom half of the sphere is irrelevant, as is the equatorial disk inside the sphere. Density at top of bump on a plane From the reasoning in Exercise 3.47, we will put image charges of ∓QR/A (which is much smaller than Q since A ≫ R) at positions ±R2 /A (which is much smaller than R), along with an image charge of −Q at y = −A. Since the two small image charges are very close to each other, they effectively form a dipole with strength p = (QR/A)(2R2 /A) = 2QR3 /A2 . From Eq. (2.36), at the top of the hemisphere the field from this dipole points downward with magnitude 2p/4πϵ0 R3 = 4Q/4πϵ0 A2 . This is twice as strong as the sum of the downward fields due to the ±Q charges (one real and one image), which is essentially equal to 2Q/4πϵ0 A2 at the top of the hemisphere. The total downward field is therefore 6Q/4πϵ0 A2 , which is three times as strong as the field in the case where we simply have a flat plane (because then only the ±Q charges are relevant). Since the electric field is zero below the conductor in any case, the surface density is proportional to the field. The surface density at the top of the hemisphere is therefore three times as large as the density in the case of the flat plane. Using the above dipole approximation, you can also show that the field (and hence density) is zero at the corner where the hemisphere meets the plane. This is consistent with the fact that the hemisphere and plane form an equipotential surface, which the electric field must be perpendicular to at all points. And the only vector that is perpendicular to two directions is the zero vector. Positive or negative density Let the point charge Q be at radius nR, where n is a numerical factor. (Working with this factor n makes the solution a little cleaner than it otherwise might be.) From Problem 3.13 there is an image charge −Q/n located at radius R/n. And then from Problem 3.16 there is an additional image charge (1 + 1/n)Q located at the center of the shell, to make the total charge on the shell be Q. In the cutoff case where the surface charge density σ at the closest point on the shell changes from negative to positive, σ will be zero. But the field right outside the shell equals σ/ϵ0 , so σ will be zero if the field is zero. The field equals the sum of the fields from the three charges (one real and two image). Being careful with the signs of the three fields, if we set the total field right outside the closest point equal to zero, we obtain (ignoring the 1/4πϵ0 ) Q/n Q (1 + 1/n)Q = + R2 (1 − 1/n)2 R2 (n − 1)2 R2 n+1 n 1 = + n (n − 1)2 (n − 1)2 √ 1 3+ 5 1 = , (249) =⇒ n2 − 3n + 1 = 0 =⇒ n = =⇒ 2 n 2 (n − 1) √ as desired. The other solution, n = (3 − 5)/2, is smaller than 1 and hence not applicable, since we are assuming n > 1 (the given charge is outside the sphere). =⇒ Remark: We can also ask the analogous question where we put a charge Q inside a nongrounded conducting spherical shell with radius R and net charge Q. A continuity argument again shows that there must be a cutoff radius, below which the charge density (the sum of the inner and outer surface densities on the shell) is positive everywhere on the shell. But the solution is slightly different, due to the fact that charge resides on both the inner and outer surfaces of the shell. The inner surface has a total charge −Q, nonuniformly distributed; the density is determined by considering an image charge located outside (see Problem 3.13). There is also a charge 2Q on the outer surface (to make the total charge on the shell equal to Q); this charge is uniformly distributed. You can show that the inner and outer densities cancel at the nearest point on the shell if the given charge Q is located at √ radius R(5 − 17)/4 ≈ (0.219)R. Attractive or repulsive? Let us write r as nR, where n is a numerical factor. (Working with this factor n makes the solution a little cleaner than it otherwise might be.) Our goal is to find the value of n for which the force on the point charge Q is zero. From Problem 3.13 there is an image charge −Q/n located at radius R/n. And then from Problem 3.16 there is an additional image charge of (1 + 1/n)Q located at the center of the shell, to make the total charge on the shell be Q. The net field at the location of the given charge Q from these two image charges is (ignoring the 1/4πϵ0 ) E= (1 + 1/n)Q (−Q/n) . + (n − 1/n)2 R2 (nR)2 (250) Setting this equal to zero yields 1/n 1 + 1/n = . n2 (n − 1/n)2 (251) Simplifying gives n+1= n4 (n2 − 1)2 =⇒ n5 − 2n3 − 2n2 + n + 1 = 0. (252) This equation factors into (253) (n2 − n − 1)(n3 + n2 − 1) = 0, √ and the roots of the quadratic are n = (1 ± 5)/2. We are concerned with the “+” root, since r > R. √ (Also, the real root of the cubic equation is less than 1.) So r/R = n = (1 + 5)/2 ≈ 1.618, as desired. Another point of interest is the location of the maximum repulsive force. You can show numerically that this location is r ≈ (2.074)R. Conducting sphere in a uniform field (a) From the reasoning in the solution to Problem 1.27, the field due to the upper sphere, at a point at position r1 with respect to its center, is E = ρr1 /3ϵ0 . Likewise, the field due to the lower (negative) sphere, at a point at position r2 with respect to its center, is E = −ρr2 /3ϵ0 . The sum of these fields is (with the subscript “s” standing for “spheres”) Es = ρs ρr1 ρr2 ρ(r1 − r2 ) =− , = − 3ϵ0 3ϵ0 3ϵ0 3ϵ0 (254) where s is the upward pointing vector between the centers. This result is independent of the position inside the cavity, so the field points downward with the uniform value of ρs/3ϵ0 throughout the overlap region. (b) The (radial) distance between the dashed circle and the bottom circle is s. And the radial displacement between the bottom circle and the top circle is s cos θ (which is negative for θ > π/2, where θ is measured down from the top of the circles), because the vertical distance between the two circles is always s, and the radial component of this distance brings in a factor of cos θ. So the total thickness of the shaded region (that is, the radial distance between the dashed circle and the top circle) is ℓ = s(1 + cos θ). This correctly equals 2s at the top of the circles and 0 at the bottom. Consider the part of the shaded region that lies in a horizontal ring (that is, one that extends into and out of the page), all of whose points subtend an angle θ with respect to the vertical. Let the tangential thickness of the ring subtend an angle dθ. The volume of the shaded region corresponding to this ring is (2πR sin θ)(R dθ)ℓ = (2πR sin θ)(R dθ)s(1 + cos θ). 2 (255) The charge in this ring is therefore ρ2πR s sin θ(1 + cos θ) dθ. Since the ring is essentially located right on the surface of the bottom sphere (if s is small), each little piece dq feels a force with magnitude Q dq/4πϵ0 R2 due to the bottom sphere, where −Q is the charge of the sphere (which equals 4πR3 ρ/3, but we won’t need this). Only the vertical component survives, and this brings in a factor of cos θ. So the vertical force due to the bottom sphere on the ring is directed downward with magnitude ( ) Q ρ2πR2 s sin θ(1 + cos θ) dθ Qρs sin θ cos θ(1 + cos θ) dθ . (256) · cos θ = 2ϵ0 4πϵ0 R2 Integrating this from 0 to π gives the total downward force on the shaded region as ( ) π ∫ π cos2 θ cos3 θ  Qρs Qρs sin θ cos θ(1 + cos θ) dθ = − −   2ϵ0 0 2ϵ0 2 3 0 = Qρs 2 Qρs · = , 2ϵ0 3 3ϵ0 (257) where Q = 4πR3 ρ/3. (c) If the field points upward, then the force on the upper (positive) sphere is simply EQ pointing upward. This is equal to the downward force we found in part (b) if 3ϵ0 E Qρs =⇒ s = EQ = . (258) 3ϵ0 ρ This distance will be small compared with R if ρ is sufficiently large. For the purposes of this problem we will assume this is the case. (d) The total field in the overlap region is the uniform field E plus the field due to the two spheres, which we found in part (a). This field points downward with magnitude ρs/3ϵ0 . Using the s we found in part (c), this equals ρ(3ϵ0 E/ρ)/3ϵ0 = E. This downward field therefore exactly cancels the upward uniform field E , so the total field is zero everywhere in the overlap region. And for small s this region is essentially the same as the interior of the spheres. (e) As we saw above in part (b), the thickness of the thin non-overlap regions in Fig. 3.34 is s cos θ. So the effective surface charge density is σ = ρ(s cos θ) = ρ 3ϵ0 E cos θ = 3ϵ0 E cos θ, ρ (259) as desired. (Note that the relevant thickness here is not the thickness of the shaded region in Fig. 3.35.) This σ = 3ϵ0 E cos θ result implies that the magnitude of the field at the poles is three times the uniform field E. The conducting-sphere limit is obtained in the ρ → ∞ and s → 0 limit, with the product ρs equalling 3ϵ0 E. Aluminum capacitor The capacitance is C= ϵ0 A = s ( 8.85 · 10−12 ) ) s2 C 2 ( π(0.075 m)2 3 kg m = 3.910 · 10−9 F = 3910 pF. (260) 4 · 10−5 m A Q s Inserting a plate Put charges Q and −Q on the two conductors in each of the two given capacitors. In the bottom capacitor in Fig. 78, one of the conductors consists of the two outer plates, because they are connected by a wire. The charge distributions on the various surfaces are shown. All the factors of 1/2 arise from symmetry. In the bottom capacitor, the potential difference (which is the difference between either of the outside plates and the inner plate) equals the field times the separation. The field is half of what it is in the top capacitor (because the density σ is half), and the separation is also half. So -Q A s/2 s/2 Q/2 Q/2 -Q/2 -Q/2 Figure 78 wire the potential difference is (1/2)(1/2) = 1/4 of what it is in the top capacitor. Since the charge Q on each capacitor is the same, we have Q = Ctopϕ, and Q = Cbottom (ϕ/4). (261) These quickly give Cbottom = 4Ctop . So our answer is 4C . In the more general case where the middle plate is a fraction f of the distance from one of the outside plates to the other, you can show that the capacitance is C/[f (1 − f )]. This correctly equals 4C when f = 1/2. It minimum when f = 1/2 and goes to infinity as f goes to 0 or 1. Dividing the surface charge If σ1 is the surface density on the top face of the inner plate, and if σ2 is the density on the bottom face, then the magnitudes of the electric fields in the top and bottom regions are E1 = σ1 /ϵ0 and E2 = σ2 /ϵ0 . These follow from using Gauss’s law with surfaces that pass through the interior of the middle plate where the field is zero. The difference in potential between the middle and top plates is E1 (0.05 m), and the difference in potential between the middle and bottom plates is E2 (0.08 m). Since the top and bottom plates are at the same potential, we must have 5E1 = 8E2 =⇒ 5σ1 = 8σ2 . Combining this with the given fact that σ1 + σ2 = σ, we quickly find σ1 = (8/13)σ and σ2 = (5/13)σ. Remark: From similar reasoning involving Gaussian surfaces with one side lying inside a conductor, it follows that the density on the bottom face of the top plate is −σ1 , and the density on the top face of the bottom plate is −σ2 . Assuming that there is zero net charge on the outer two plates, this leaves at total of σ1 + σ2 = σ for the outer surfaces of these plates. It must get divided evenly, because otherwise these two surfaces would create a nonzero field between them, which would change the above fields and make the outer plates not be at the same potential. If any additional charge is dumped on the outer plates, it simply gets divided evenly between their two outer surfaces. Two pairs of plates wire E= 0 Q1 wire E= 0 Q2 Figure 79 Since the top two plates are at the same potential, the field is zero between them. Likewise, the field is zero between the bottom two plates. This exercise is therefore basically the same as Problem 3.20, due to the fact that the field inside a conductor is zero, just as the field between the two pairs of plates is zero in the present exercise. The four sheets here are equivalent to the four surfaces of the two plates in Problem 3.20. The solution is therefore basically the same. Consider the Gaussian surface indicated by the dotted box in Fig. 79. Since there is no flux out of the top or bottom, the net charge enclosed must be zero. Hence there are equal and opposite charges on the inner two plates. We now claim that the charges on the outer two plates must be equal. This is true because the two inner plates create zero net field in the E = 0 regions (because these two plates are on the same side of each of the E = 0 regions and have opposite charges, so their fields cancel). The outer two sheets must therefore also create zero net field in the E = 0 regions. Since these sheets are on opposite sides of a given E = 0 region, their charges must be the same if the fields are to cancel. We can therefore describe the charges on the four plates, from top to bottom, as q1 , q2 , −q2 , and q1 . The given information tells us that q1 + q2 = Q1 and q1 − q2 = Q2 . Solving for q1 and q2 gives q1 = (Q1 + Q2 )/2 and q2 = (Q1 − Q2 )/2. So from top to bottom, the charges on the four plates are Q1 + Q2 Q 2 − Q1 Q1 + Q 2 Q1 − Q2 , , , 2 2 2 2 (262) We can also state which four of the eight surfaces (top and bottom of the each of the four plates) these charges lie on. None of the charges can border the E = 0 regions, because otherwise the standard argument involving a Gaussian surface with one face lying inside the metal of the conductor would imply a nonzero field in these regions. So the four charges lie on the top of the top plate, the bottom of the next, the top of the next, and finally the bottom of the bottom plate. Field just outside a capacitor If the disks were infinitely large, the desired field would be zero. But with a finite R, the repulsive field from the positive disk (which acts like an infinite plane, for points infinitesimally close to it) is slightly larger than the attractive field from the negative disk, which does’t quite act like an infinite plane. Let’s find the field due to a disk with radius R and surface density σ, at a general point a distance z from the center of the disk along the axis. This can be found by slicing up the disk into rings and finding the z component of the field due to the charge in √ each ring. We obtain (the z/ r2 + z 2 factor here gives the z component): E = = ∫ σz R z r dr (2πr dr)σ = ·√ 2 2 2 2 2ϵ0 0 (r2 + z 2 )3/2 r +z r +z 0 R  σz σz  = σ − √ . − √ 2 2ϵ0 2ϵ0 R2 + z 2 2ϵ0 r + z 2 0 1 4πϵ0 ∫ R (263) As expected, if z ≪ R we obtain the standard σ/2ϵ0 field from an infinite plane. In the case of the negative disk in this problem, z equals the separation s. So the difference in the magnitudes of the (oppositely √ pointing) fields from the two disks, at a point just outside the positive disk, is σs/2ϵ0 R2 + s2 . The net field therefore has this magnitude and is directed away from the positive disk. In the (usual) case at hand where s ≪ R, the net field is essentially equal to σs/2ϵ0 R, which is s/R times the σ/2ϵ0 field from an infinite plane. A 2N -plate capacitor This solution requires only a slight modification of the solution to Problem 3.21. Let the charge densities on the first N plates be σ1 , −σ2 , σ3 ,. . .. Then by left-right symmetry, the charge densities on the second N plates are . . .,−σ3 , σ2 , −σ1 . The total charge is zero, so there is no field outside the plates. Hence the field between the 1st and 2nd plates is σ1 /ϵ0 . The potential difference between these plates is therefore ϕ = σ1 s/ϵ0 . The magnitude of the potential difference is the same between all pairs of adjacent plates, because all of the odd-numbered plates have the same potential due to the connecting wires, as do all of the even-numbered plates. So the field between any two adjacent plates is σ1 /ϵ0 , with the direction alternating as shown in Fig. 80. A Gaussian surface spanning any of the interior plates tells us that all of these plates have charge density ±2σ1 . The total charge on the N positive plates is therefore Q = (σ1 + (N − 1)2σ1 )A, which gives σ1 = Q/(2N − 1)A. The potential difference σ1 -σ2 σ3 -σ4 σ __1 ε0 σ5 ... wires Figure 80 between the positive and negative (sets of N ) plates in the capacitor can then be written as ) ( (2N − 1)Aϵ0 σ1 s Qs (2N − 1)Aϵ0 . ϕ= = ϕ =⇒ C = =⇒ Q = s s ϵ0 (2N − 1)Aϵ0 (264) As N → ∞, this is essentially equal to 2NAϵ0 /s. The capacitance in Eq. (264) is larger than the capacitance we would obtain if we juxtapose the two pairs of N plates to create two plates with area N A. If we keep the separation s, then the capacitance of the resulting standard two-plate capacitor would be C = ϵ0 (N A)/s. As mentioned in the solution to Problem 3.21, the reason why the capacitance in Eq. (264) is larger than C = ϵ0 (N A)/s is because we effectively have 2N − 1 identical area-A capacitors of alternating orientation lined up next to each other, instead of N area-A capacitors (with the same orientation). Capacitor paradox The second reasoning is correct. Plate 3 is indeed at a lower potential than plate 1, so charge will flow. The error in the first reasoning is encompassed in the word, “So.” Although it is true that the potential differences of the two capacitors are the same, this does not imply that the potentials of the two positive plates are equal. If we arbitrarily assign zero potential to plate 1, and if the common potential difference is ϕ, then the potentials of the four plates are, from left to right, 0, −ϕ, −ϕ, and −2ϕ. No matter where we define the zero of potential, the potential of the leftmost plate is ϕ larger than the potential of the third plate, and 2ϕ larger than the potential of the rightmost plate. Coaxial capacitor Neglecting end effects, we can assume that the charge ±Q is uniformly distributed along each cylinder. The field between the cylinders is that of a line charge with density λ = Q/L, so E = λ/2πϵ0 r = Q/2πϵ0 Lr. The magnitude of the potential difference between the cylinders is then ∫ a ∫ a (a ) Q Q dr |∆ϕ| = E dr = = ln . (265) Lr L 2πϵ 2πϵ b 0 0 b b Since C = Q/|∆ϕ|, the capacitance is given by C = 2πϵ0 L/ ln(a/b). If a − b ≪ b, then we can use the Taylor series ln(1 + ϵ) ≈ ϵ to write ( ) (a ) a−b a−b ln = ln 1 + ≈ . (266) b b b So the capacitance becomes C ≈ 2πϵ0 bL/(a − b). But 2πbL is the area A of the inner cylinder, and a − b is the separation s between the cylinders. So the capacitance can be written as C = ϵ0 A/s, which agrees with the standard result for the parallel-plate capacitor. A three-shell capacitor (a) Let Q1 and Q3 be the final charges on the inner and outer shells, respectively. The outward-pointing field between the inner and middle shells is due only to the inner shell, and it equals (ignoring the 1/4πϵ0 since it will cancel) Q1 /r2 . So the potential difference between the inner and middle shells is Q1 (1/R − 1/2R), with the inner shell at the higher potential. If the inner and outer shells are at the same potential, then Q1 (1/R − 1/2R) must also be the potential difference between the outer and middle shells, with the outer shell at the higher potential. The field between the middle and outer shells must therefore point inward. This field is due to the inner two shells, so it points inward with magnitude (Q − Q1 )/r2 , given that −Q is the charge on the middle shell. Note that Q must be larger than Q1 . The potential difference between the outer two shells is then (Q − Q1 )(1/2R − 1/3R), with the outer shell at the higher potential. Equating the inner-middle and outer-middle potential differences gives ( ) ) ( 1 Q 1 1 Q − Q1 1 Q1 Q1 =⇒ Q1 = . − − = =⇒ = (Q − Q1 ) 6 4 2R 3R R 2 2R (267) And then Q3 = 3Q/4, to make the total charge on the inner and outer shells be equal to Q. (b) The potential difference between the inner and middle shells, which is the same as the difference between the outer and middle shells, is (bringing the 1/4πϵ0 back in, and using Q1 = Q/4) ) ( 1 Q Q1 1 − . (268) ϕ= = 32πϵ0 R 2R 4πϵ0 R Therefore Q = (32πϵ0 R)ϕ, so the capacitance is 32πϵ0 R. (c) Note that Q3 didn’t appear anywhere in the calculation in part (a). It can therefore take on any value, and the inner-middle and outer-middle potential differences will still be equal, provided that Q1 = Q/4. So if we add charge q to the outer shell, it will simply stay there, uniformly distributed on the outside surface of the shell. It will raise the potential everywhere inside by q/4πϵ0 (3R), but since this change is uniform inside, all differences remain the same. If any charge flowed across the wire from the outer shell to the inner shell, the final charge Q1 on the inner shell would violate the Q1 = Q/4 result we found above (because the middle shell is now isolated, so its charge of −Q doesn’t change). Capacitance of a spheroid If b ≈ a, then ϵ ≪ 1, and we can use the Taylor approximation, ln(1 ± ϵ) ≈ ±ϵ. The capacitance then becomes C= 8πϵ0 aϵ 8πϵ0 aϵ ≈ = 4πϵ0 a, 2ϵ ln(1 + ϵ) − ln(1 − ϵ) (269) in agreement with the capacitance of a sphere given in Eq. (3.10). The stored energy equals Q2 /2C. Since Q is held constant, we need only determine how C changes as the sphere is deformed. We will find that C increases, which means that the stored energy decreases. Let C0 be the capacitance of a sphere of unit radius, a = b = 1. And let C be the √ capacitance of a prolate spheroid of equal volume. This spheroid has b = 1/ a, since the volume equals (4/3)πab2 . (We’ll ignore the units of a and b. Equivalently, a and b are defined as the dimensionless ratios of the new lengths to the original radius of the sphere.) We have √ C 2aϵ 1 ( (270) ), = where ϵ = 1 − 3 . a 1+ϵ C0 ln 1−ϵ C/C0 1.5 1.0 0.5 a 0.0 0 1 2 3 4 Figure 81 b a Figure 82 In the limit where a is very large (so we have a long stick-like object), you can show with a Taylor series that C/C0 ≈ 2a/ ln(4a3 ) ≈ 2a/(3 ln a), to leading order. This grows with a, so a long spheroid-shaped stick can have a capacitance much greater than that of a sphere of equal volume. If you want to write things more generally in terms of both a and b, the capacitance given in the problem equals C ≈ 4πϵ0 a/ ln(2a/b) in the a ≫ b limit. As an example, consider a spheroid with a = 1 km and b = 1 mm. Its volume is that of a sphere with radius (ab2 )1/3 = 0.1 m, but its capacitance is that of a sphere with radius r = 69 meters, because C ≈ 4πϵ0(10 m)3 / ln(2 · 106 ) = 4πϵ0 (69 m). Deriving C for a spheroid h stick A plot of C/C0 is shown in Fig. 81. (The expression for the capacitance given in the statement of the problem is valid only for a prolate spheroid, that is, one with a > 1.) As promised, C is larger than C0 . a We’ll assume here the validity of the result from Exercise 2.44, namely that the potential due to a stick with uniform charge density is constant over an ellipsoid that has the ends of the stick as its foci. Such an ellipse is shown in Fig. 82. The axes have lengths 2a√and 2b, and the stick has length 2h. From the properties of an ellipse, we have a = h2 + b2 . To find the (constant) potential over this ellipse, we may conveniently pick points at the ends of either axis. An end of the major axis yields a slightly simpler integral (you can check that an end of the minor axis yields the same result). We have ( ) √ ( ) ∫ h a+h a + a 2 − b2 λ λ λ dy 1 √ ln ln = ϕ= = 4πϵ0 4πϵ0 a−h 4πϵ0 −h a − y a − a 2 − b2 ( ) 1+ϵ λ ln = , (271) 4πϵ0 1−ϵ where √ b2 . a2 The charge contained within the ellipsoid is the charge on the stick, √ Q = (2h)λ = 2λ a2 − b2 = 2λaϵ. ϵ≡ 1− (272) (273) From the reasoning near the end of Section 3.4, in the region of space exterior to the conducting ellipsoid, the field due to the ellipsoid with charge Q is identical to the field due to the uniform stick with charge Q. This is true because the latter field satisfies the boundary conditions for the ellipsoid (the field is perpendicular to the surface, and the total flux is Q/ϵ0 ), so the uniqueness theorem tells us that this solution must be the solution for the ellipsoid. Using Q = 2λaϵ in the Q = Cϕ relation for the ellipsoid gives C= as desired. Q = ϕ 2λaϵ 8πϵ0 aϵ )= ), ( ( λ 1+ϵ 1+ϵ ln ln 1−ϵ 1−ϵ 4πϵ0 (274)