Electrostatics Part 2, Solution Manual

Uniform field strength Let Qr be the charge inside radius r. Since Gauss’s law tell us that Er ∝ Qr /r2 , we therefore want Qr ∝ r2 if Er is to be constant. That is, we want ∫ r 4πx2 ρ(x) dx ∝ r2 , (65) 0 for any value of r up to the radius∫of the sphere. We quickly see that this proportionality holds if ρ(x) ∝ 1/x, because (x2 /x) dx = x2 /2. So this is the desired form of ρ; it is inversely proportional to the radius. Although ρ diverges at the origin, the charge there is still finite (the amount of charge within a sphere with radius r is proportional to r2 ) because of the 4πx2 in the volume element in the above integral. Note that the field right at the center isn’t well defined. Alternatively (or rather, equivalently), since Gauss’s law tells us that we want Qr = Br 2 , for some constant B, we have dQr /dr = 2Br. But a general expression for dQr /dr is 4πr2 ρ (if we imagine adding on a thin shell). Equating these two expressions for dQr /dr gives ρ = B/2πr ∝ 1/r, as desired. The solution for the case of a cylinder is similar. Again let Qr,ℓ be the charge inside radius r, for a length ℓ of the cylinder. Since Gauss’s law tell us that Er ∝ Qr,ℓ/rℓ, we therefore want Qr,ℓ ∝ rℓ if Er is to be constant. That is, we want ∫ r 2πxℓρ(x) dx ∝ rℓ, (66) 0 for any value of r up ∫ to the radius of the cylinder. This proportionality holds if ρ(x) ∝ 1/x, because (x/x) dx = x. So the answer is the same as for the sphere. Alternatively, Gauss’s law tells us that we want Qr,ℓ = Brℓ, for some constant B, so dQr,ℓ/dr = Bℓ. But dQr,ℓ/dr also equals 2πrℓρ (if we imagine adding on a thin shell). Equating these two expressions for dQr,ℓ/dr gives ρ = B/2πr ∝ 1/r, as desired. Interestingly, the answer changes if we kick the dimension down one more step and consider a slab. Since the field due to a charged sheet is uniform, we know that the magnitude of the field will be independent of position inside the slab if the slab has ρ = 0 everywhere except for a sheet of charge at its center plane. Mathematically, let Qr,A be the charge inside a sub-slab with area A and thickness 2r centered with respect to the given slab. Gauss’s law tells us that we want Qr,A = BA for some constant B, so dQr,A /dr = 0. But dQr,A/dr also equals 2Aρ (if we imagine adding on two thin surfaces on the two faces of the sub-slab). Equating these two expressions for dQr,A/dr gives ρ = 0. This holds everywhere except at the center plane at r = 0. The reasoning in the above three cases (indexed by n = 2, 1, 0, respectively) can be concisely summarized as follows: Qr = Brn =⇒ dQr /dr = nBrn−1 . But also dQr /dr = krn ρ. Therefore ρ ∝ n/r. The n in the numerator makes it clear why the n = 0 case is different from the others. Carved-out sphere The given setup is equivalent to the superposition of a sphere with radius a and density ρ, plus an off-center sphere with radius a/2 and density −ρ. The desired fields at A and B are the sums of the fields from these two objects. The charge in the big sphere is Qb = (4/3)πa3 ρ, while the charge in the small sphere is Qs = (4/3)π(a/2)3 (−ρ) = −Qb /8. For convenience, let the field at B due to the big sphere be labeled E0 . Then E0 ≡ Eb,B = (4/3)πa3 ρ Qb aρ , = = 4πϵ0 a2 3ϵ0 4πϵ0 a2 (67) and the field is directed downward. The field at A due to the big sphere is Eb,A = 0. The field at A due to the small (negative) sphere has magnitude Es,A = aρ (4/3)π(a/2)3 ρ Qs E0 = , = = 2 4πϵ0 (a/2)2 4πϵ0 (a/2)2 6ϵ0 (68) and is directed upward. The field at B due to the small sphere has magnitude Es,B = aρ (4/3)π(a/2)3 ρ Qs E0 = , = = 18 4πϵ0 (3a/2)2 4πϵ0 (3a/2)2 54ϵ0 (69) and is directed upward. 3σ0 σ0 ___ 2ε0 The total field at A is therefore directed upward with magnitude 0 + E0 /2 = aρ/6ϵ0 . And the total field at B is directed downward with magnitude E0 −E0 /18 = 17E0 /18 = 17aρ/54ϵ0 . -2σ0 5σ0 ___ 2ε0 Field from two sheets The field from an infinite sheet with charge density σ has magnitude σ/2ϵ0 . It is directed away from the sheet if σ is positive, and toward it if σ is negative. The total field in the given setup equals the superposition of the fields from each sheet; the result is shown in Fig. 23. The field has magnitude (3σ0 + 2σ0 )/2ϵ0 = 5σ0 /2ϵ0 inside the sheets and (3σ0 − 2σ0 )/2ϵ0 = σ0 /2ϵ0 outside the sheets. In all regions it is directed away from the 3σ0 sheet. σ0 ___ 2ε0 Figure 23 -2σ0 3σ1 ___ 2ε0 2σ ___1 2ε0 If the sheets intersect at right angles, the field is again obtained by superposition, but now the two individual fields are orthogonal. Fig.√24 shows the results in the four regions. The magnitude of the field everywhere is 32 + 22 (σ0 /2ϵ0 ) ≈ (1.8)σ0 /ϵ0 . In all regions it is directed at least partially away from the 3σ0 sheet and partially toward the −2σ0 sheet. Intersecting sheets 3σ0 (a) The electric field from a given sheet points away from the sheet and has uniform magnitude σ/2ϵ0 . The three fields at a given point therefore take the form shown in Fig. 25. At the location shown, the net field is directed exactly rightward and has magnitude E = E2 + (E1 + E3 ) cos 60◦ = σ σ E3 60 1 3 2 Figure 25 (70) The field has the same magnitude and direction (to the right) everywhere in the rightmost “pie piece,” because the field due to a sheet doesn’t depend on the distance from the sheet. Similarly, in the other five pie pieces the magnitude is σ/2ϵ0 , and the direction is parallel to the line that bisects the angle of the pie piece. Figure 24 σ σ σ σ +2 cos 60◦ = . 2ϵ0 2ϵ0 ϵ0 E1 E2 σ σ (b) Fig. 26 shows the field at points on a symmetrically-located hexagon. Let the “radius” of the hexagon be r, and consider a hexagonal tube with length ℓ perpendicular to the page. The surface area of this tube is 6rℓ, and the charge enclosed is 6rℓσ. Since the electric field is everywhere perpendicular to the surface, Gauss’s law gives ∫ 6rℓσ σ Q =⇒ E · 6rℓ = =⇒ E = , (71) E · da = ϵ0 ϵ0 ϵ0 σ r r r Figure 26 in agreement with the result in part (a). Again, note that E is independent of r. While Gauss’s law is always valid, it was actually useful in the present setup because we were able to find a simple surface that is everywhere perpendicular to the electric field (because the electric field is uniform in each pie piece). (c) For general N , the electric field is everywhere perpendicular to a regular 2N -gon. ( ) The surface area of this 2N -gon is (2N ) 2 sin(π/2N ) rℓ, and the charge enclosed is (2N )rℓσ. So Gauss’s law gives ( ) (2N )rℓσ E · (2N ) 2 sin(π/2N ) rℓ = ϵ0 =⇒ E = σ . 2ϵ0 sin(π/2N ) (72) As expected, this is independent of r. And it agrees with the above result when N = 3. For large N , we have sin(π/2N ) ≈ π/2N , so E ≈ N σ/πϵ0 . In the case of large N , the sheets are very close to each other, so we effectively have a continuous volume charge distribution that depends on r. The separation between adjacent sheets grows linearly with r, so we have ρ(r) ∝ 1/r. More precisely, you can show that ρ(r) = N σ/πr. This is consistent with the result from Exercise 1.68, where we found that a cylinder with a density of the form ρ(r) ∝ 1/r produces a field whose magnitude is independent of r (inside the cylinder). A plane and a slab The total effective charge per unit area (looking perpendicular to the sheet/slab) is σ+ρd, because ρ(Ad) is the charge contained within an area A of the slab. Let x = 0 be defined to be the location of the plane. Then for x < 0 the field is E = −(σ + ρd)/2ϵ0 , and for x > d it is E = (σ + ρd)/2ϵ0 . At a general point inside the slab (that is, for 0 < x ≤ d), there is a charge density σ + ρx to the left of the point and (d − x)ρ to the right. So for 0 < x ≤ d the field is σ + ρx σ − ρd + 2ρx (d − x)ρ . = − 2ϵ0 2ϵ0 2ϵ0 σ (73) ρ The plot of E as a function of x is shown in Fig. 27. E is continuous at x = d but not at x = 0. If the plane had a nonzero thickness, then the field would be continuous at x = 0. The case shown in the plot has ρd > σ. If we instead had σ > ρd, then at the discontinuity at x = 0, E would jump to a positive value. Sphere in a cylinder From the reasoning in the solution to Problem 1.27, the electric field inside a uniform cylinder is E = ρr/2ϵ0 , where r points away from the axis. And the electric field inside a uniform sphere is E = ρr/3ϵ0 , where r points away from the center. The given setup may be considered to be the superposition of a uniform cylinder with density ρ and a uniform sphere with density −3ρ/2. This produces the desired net density of −ρ/2 within the sphere. x x =d Ex σ+ρd _____ 2ε0 ρd __ d ε0 σ+ρd - _____ 2ε0 σ __ ε0 Figure 27 x Consider a point inside the sphere. Let its position with respect to the axis of the cylinder be rc , and let its position with respect to the center of the sphere be rs . Then using the above forms of the fields, along with superposition, we find that the field at this point is ρrc (−3ρ/2)rs ρ E = E c + Es = = + (rc − rs ). (74) 2ϵ0 3ϵ0 2ϵ0 y rc rs rc - rs = ax Figure 28 x Let’s look at this vector rc − rs . If the given point lies in the x-y plane, then we have the situation shown in Fig. 28; the axis of the cylinder (the z axis) points out of the page. The difference rc − rs is simply the vector ax̂. If the given point does not lie in the x-y plane, then rc will still be parallel to the x-y plane, but rs will now have a z component. However, this z component doesn’t affect the x-y component of the difference rc − rs , so the x-y component still equals ax̂. In short, a “top” view of the setup (looking along the z axis) makes it clear that the projection of rc − rs onto the x-y plane always equals ax̂. From Eq. (74), the x-y component of the field inside the sphere therefore equals (ρa/2ϵ0 )x̂, which is uniform, as desired. In the special case where the sphere is centered on the z axis (so that a = 0), there is additionally zero x component, so the field points only in the z direction inside the sphere, with a magnitude dependent only on z. This dependence is linear, so a charge will undergo simple harmonic motion if its initial velocity is in the z direction. Zero field in a sphere ∫ Gauss’s law says S E · da = Q/ϵ0 , where Q is the charge enclosed by the surface S . If we draw a spherical Gaussian surface with radius r inside the given sphere, Gauss’s law gives ρr (4πr 3 /3)ρ =⇒ E = . (75) E · 4πr2 = ϵ0 3ϵ0 The full vector form of this field is E = r̂ρr/3ϵ0 . But rr̂ is simply the vector (x, y, z), so Esphere = (ρ/3ϵ0 )(x, y, z). Similarly, if we draw a cylindrical Gaussian surface with radius r and length ℓ inside the given cylinder, Gauss’s law gives E · 2πrℓ = (πr 2 ℓ)ρ ϵ0 =⇒ E = ρr . 2ϵ0 (76) The full vector form of this field is E = r̂ρr/2ϵ0 , where the ˆr vector here represents the direction away from the z axis. So rr̂ is the vector (x, y, 0). Hence Ecyl = (ρ/2ϵ0 )(x, y, 0). Finally, if we draw a slab Gaussian surface with area A and thickness 2z, inside the slab and centered in the slab, Gauss’s law gives E · 2A = (2zA)ρ ϵ0 =⇒ E = ρz . ϵ0 (77) This field points in the z direction, so it is just Eslab = (ρ/ϵ0 )(0, 0, z). If we now give the three objects the densities ρ1 , ρ2 , ρ3 , we find that the total field at a given point (x, y, z) inside all three objects (which means inside the sphere) equals Etotal = ) 1 ( ρ1 ρ2 (x, y, z) + (x, y, 0) + ρ3 (0, 0, z ) . 2 ϵ0 3 (78) We want this to be zero. The x and y components will equal zero if 3 ρ1 ρ 2 + = 0 =⇒ ρ1 = − ρ2 . 2 2 3 (79) And the z component will equal zero if ρ1 + ρ3 = 0 =⇒ ρ1 = −3ρ3 . 3 (80) These equations will be satisfied if the densities are in the ratio of ρ1 : ρ2 : ρ3 = 3 : −2 : −1. As a double check, the sum of these three densities (which is the net density inside the sphere) equals zero. This is correct, because if the field inside the sphere is zero, then any volume we choose inside the sphere must contain zero charge, because there is zero flux through its surface. The only way this can happen is if there is no charge anywhere inside the sphere. Hence ρ = 0. This is consistent with the differential form of Gauss’s law, ∇ · E = ρ/ϵ0 , which we’ll get to in Chapter 2; E = 0 implies ρ = 0. Ball in a sphere (a) At a radius r inside a sphere with charge density ρ, the electric field is effectively due to the charge inside radius r. So the field is E= (4πr3 /3)ρ ρr . = 3ϵ0 4πϵ0 r2 (81) The field points radially outward (for positive ρ), so we can write the E vector compactly as E = ρr/3ϵ0 . The force on the smaller ball due to the larger sphere is found by integrating the effect of the E field on all the pieces of the ball. This integral is easy to do if we group the pieces in pairs that are symmetrically located on either side of the center of the ball (at position a). Consider two pieces located at positions a + r′ and a − r′ . Then because the above E field is linear in r, the r′ parts of the sum cancel, and we end up with two pieces effectively located at the center of the ball. We can build up the entire ball from such pairs, so we see that the ball can be effectively treated like a point charge at its center, as far as the force from the larger sphere goes. The force is therefore the same as if it were a point charge of the same charge q . It’s no surprise that the ball acts effectively like a point charge at its center, because what we just did is basically find the average value of r over the ball, which is the same thing we do when we find the center of mass of a uniform object. And the CM of a sphere is at its center. So more generally, if we have an oddly-shaped charged object instead of a nice ball (and even if it isn’t uniform), and if it is located entirely within a uniform sphere of charge, then the force on it is the same as if all of its charge were located at its “center of charge.” (b) Now consider the slightly different setup where we remove the charge in the larger sphere where the ball is. This is a more realistic scenario, because it corresponds to hollowing out a cavity and then filling it up with another material. It turns out that the force on the ball doesn’t change. This follows from the fact that the larger-sphere-with-cavity is the superposition of the complete larger sphere plus a sphere of negative charge density located where the ball is. So the total force on the ball is the sum of the forces from the original complete sphere plus the new sphere of negative charge. But the latter provides no force on the ball, by symmetry, because it is located in exactly the same place as the ball. The total force is therefore the same as before. The result of this exercise is consistent with the result in Problem 1.27, which tells us that the electric field is uniform inside a spherical cavity within a uniform sphere. Hydrogen atom The fraction of the negative charge that lies inside a sphere of radius r1 is ∫ r1 ∫ r1 −2r/a ∫ r1 2 −2r/a ∫ x1 2 −x 0 0 Ce 4πr2 dr r e dr x e dx ρ dv ∫0∞ = ∫0∞ −2r/a = ∫ 0∞ 2 −2r/a = ∫0∞ 2 −x , 2 0 0 ρ dv Ce 4πr dr r e dr x e dx 0 0 0 0 (82) where we have made the change of variables x ≡ 2r/a0 . Note that we don’t need to know C, because it cancels. You can verify the integral: ∫ x1 x1  (83) x2 e−x dx = −(x2 + 2x + 2)e−x  = 2 − (x21 + 2x1 + 2)e−x1 . 0 0 For x1 = ∞ (that is, r1 = ∞) the integral is 2. And for x1 = 2 (that is, r1 = a0 ) the integral is 2 − 10e−2 . The fraction of the full electron charge −e that lies inside radius a0 is therefore (2 − 10e−2 )/2 = 1 − 5/(2.718)2 = 0.323. The net positive charge inside r = a0 is then q = (1 − 0.323)(1.6 · 10−19 C) = 1.08 · 10−19 C. This field at this radius is ( ) kg m3 1 q 1.08 · 10−19 C = 9 · 109 2 2 ≈ 3.5 · 1011 N/C, (84) 4πϵ0 a20 (0.53 · 10−10 m)2 s C which is huge! Electron jelly The force on a proton, at radius r, from the electron jelly is effectively due to the jelly that is inside radius r. The force points toward the center of the sphere. If the net force on a proton is zero, the force from the other proton must also point along the line (away) from the center. The two protons must therefore lie on the same diameter. They also must be the same distance r from the center; this is true because they feel the same force (in magnitude) from each other, so they must also feel the same force from the jelly, which implies that they must have the same value of r. Since volume is proportional to r3 , the negative charge inside radius r equals −2e(r3 /a3 ). The field at radius r due to the jelly is therefore 2e(r3 /a3 ) er =− . (85) 2πϵ0 a3 4πϵ0 r2 ( ) The field at one proton due to the other is e/ 4πϵ0 (2r)2 . So the total field at one of the protons will equal zero if − e er = 2πϵ0 a3 4πϵ0 (2r)2 =⇒ r3 = a3 8 =⇒ r = a . 2 (86) This factor of 1/2 is reasonably clear in retrospect. If all of the −2e electron charge were located in a point charge at the center, it would provide a force on one of the protons that is 8 times the force due to the other proton (because the other proton is twice as far away and half as big). So the forces will balance if we reduce the effective electron charge by a factor of 8. This is accomplished by reducing the effective radius of the jelly by a factor of 2. hole Hole in a shell First solution: We can solve this exercise by direct integration. Let’s slice up the spherical shell (minus the hole) into rings parameterized by the angle θ shown in Fig. 29. The width of a ring is R dθ, and its circumferential length is 2π(R sin θ). So its area is 2πR2 sin θ dθ. All points on the ring are a distance 2R sin(θ/2) from the center of the hole. Only the vertical component of the field survives, and this brings in a factor of sin(θ/2), as you can check. If the edge of the hole is at the small angle θ = ϵ, the total field at the middle of the hole is (writing sin θ as 2 sin(θ/2) cos(θ/2)) 1 4πϵ0 ∫ π ϵ σ2πR2 sin θ dθ ( )2 sin(θ/2) 2R sin(θ/2) in the limit where ϵ → 0. = = σ 4ϵ0 ∫ Figure 29 π cos(θ/2) dθ ϵ π  σ σ sin(θ/2)  ≈ , 2ϵ0 2ϵ0 ϵ (87) Second solution: The given setup with the hole is the superposition of a complete spherical shell with density σ plus a small disk with density −σ. And very close to the center of the disk, the disk looks essentially like an infinite plane. The fields due to these two objects, at the point in question, are shown in Fig. 30. The sum of the fields at the center of the hole is therefore a field with magnitude σ/2ϵ0 , directed radially outward. Note that we obtain an outward σ/2ϵ0 field independent of whether we look at a point just inside or just outside the shell; these points yield 0 + σ/2ϵ0 or σ/ϵ0 − σ/2ϵ0 , respectively. In other words, even though the field isn’t continuous across the original complete shell or across the disk, it is continuous across the hole. It must be continuous, of course, because there is nothing but vacuum in the hole. E =0 inside __ σ ε0 + ___ σ 2ε0 Figure 30 Third solution: We can also solve this exercise by considering the force on the little disk, while it is still in place in the shell. If A is the area of the disk, then we know from Eq. (1.49) that the force on it is 0 + σ/ϵ0 σ E1 + E2 = Aσ = Aσ · . 2 2 2ϵ0 ___ σ ___ σ 2ε0 2ε0 negative disk Since the field inside the complete sphere is zero, the field inside the sphere-plus-hole is exactly the same as the field due to the negative disk. The field lines due to a disk are shown in Fig. 2.12. Near the edge of the hole, the tangential component of the field diverges. But at points in the hole exactly on the (removed) surface of the sphere, the radial component of the field is exactly σ/2ϵ0 , over the entire area of the hole. Aσ R θ (88) But the force on the disk equals the charge on the disk times the field at the location of the disk, due to all the other charge in the system (that is, the shell with the disk removed). Equation (88) therefore tells us that the (radial) field of the shell-minus-disk must be σ/2ϵ0 , as desired. Forces on three sheets From Eq. (1.49) the force per unit area on a sheet is (E1 + E2 )σ/2, where E1 and E2 are the electric fields on either side. The fields in the various regions can be found by the superpositions of the fields from the individual sheets, using the fact that the field due to a given sheet is σ/2ϵ0 . With σ0 ≡ 10−5 C/m2 , the fields in the two middle regions are 4σ0 /ϵ0 upward and 3σ0 /ϵ0 downward, as shown in Fig. 31. Above and below all three plates the field is zero. The forces per unit area on the three sheets E =0 σ =- 4σ0 A 4σ0 ___ ε0 B 3σ0 ___ ε0 C E =0 Figure 31 σ =7σ0 σ =- 3σ0 are therefore FA = FB = FC = ( ) 4σ0 N 1 8σ 20 1 = −90.4 2 , 0+ (−4σ0 ) = − (E 1 + E 2 )σ = 2 ϵ0 ϵ0 2 m ( ) N 1 3σ0 1 4σ0 7σ 2 0 = 39.5 2 , − (E 1 + E 2 )σ = (7σ0 ) = 2ϵ0 2 m 2 ϵ0 ϵ0 ( ) 1 3σ0 N 1 9σ 20 (E 1 + E 2 )σ = = 50.8 2 . 0− (−3σ0 ) = 2 m 2 ϵ0 2ϵ0 (89) The sum of these forces per area is (σ02 /ϵ0 )(−8 + 7/2 + 9/2). This equals zero as it must, because a system can’t exert a net force on itself (otherwise momentum wouldn’t be conserved). Alternatively, we can find the forces by calculating the field at the location of a given plate due to the other two plates. For example, the bottom two plates produce a field of (7 − 3)σ0 /2ϵ0 = 2σ0 /ϵ0 at the location of the top plate, which therefore feels a force per unit area equal to (2σ0 /ϵ0 )(−4σ0 ) = −8σ02/ϵ0 , as above. The (E1 + E2 )/2 averages in Eq. (89) are simply a way of finding the field at the location of one sheet due to the others. Force in a soap bubble The field just outside the balloon is E = Q/4πϵ0 R2 , and the field inside is zero. So the average field at the surface is E/2. The charge density is σ = Q/4πR2 . From Eq. (1.49), the force per unit area (that is, the pressure) is therefore P ≡ Q2 F E Q 1 Q = =σ = · . 32π 2 ϵ0 R4 A 2 4πR2 2 4πϵ0 R2 (90) Consider the two hemispheres defined by the horizontal great circle. The horizontal components of the forces on the different parts of the upper hemisphere cancel by symmetry, so we care only about the vertical components. The vertical component of the force on a little patch with area A is P A cos θ, where θ is the angle that the plane of the patch makes with the horizontal. If we write this as P (A cos θ), we see that the vertical force on a patch equals P times the projection of the area of the patch onto the horizontal plane. Since P is constant, and since the sum of all the projections of the patches in a hemisphere is simply the great-circle area πR2 , we find that the total upward force on the hemisphere is P · πR2 = Q2 . 32πϵ0 R2 (91) By comparison with Coulomb’s force law, this has the correct units of (charge)2 /[ϵ0 · (length)2 ]. It makes sense that it grows with Q and decreases with R. If you don’t want to use the above reasoning involving the projection, you can do an integral. Slice the sphere into rings, with θ being the angle of a ring down from the top of the sphere. The area of a ring is da = 2π(R sin θ)(R dθ), and the vertical component of the force on the ring is (P da) cos θ. Integrating from θ = 0 to θ = π/2 gives the total vertical force on the upper hemisphere as ) ∫ π/2 ( ∫ π/2 Q2 (P da) cos θ = (2πR2 sin θ dθ) cos θ 2 4 32π ϵ0 R 0 0 ∫ π/2 Q2 sin θ cos θ dθ = 16πϵ0 R2 0  2  π/2 2 Q Q2 sin θ  = = , 32πϵ R2 16πϵ R2 2  0 0 (92) 0 in agreement with the above result. Energy around a sphere The energy density is ϵ0 E 2 /2, where E = Q/4πϵ0 r2 . The field is zero inside the sphere of radius R, so the energy contained within a sphere of radius R1 is ( ) ( ) ∫ R1 ∫ R1 1 dr R Q2 1 Q2 Q2 1 ϵ0 E 2 − 4πr2 dr = 1− . (93) = = 2 R R R 8πϵ 2 8πϵ R r 8πϵ 1 0 1 0 R 0 R The total amount of energy out to infinity is obtained by setting R1 = ∞, which gives Q2 /8πϵ0 R. We therefore want the factor (1 − R/R1 ) to equal 9/10, which implies that R1 = 10R. Energy of concentric shells (a) The field is nonzero only in the region a < r < b, where it equals E = Q/4πϵ0 r2 . The total energy is therefore )2 ∫ b ) ( ( ∫ b Q2 1 ϵ0 E 2 1 ϵ0 Q 1 4πr2 dr = U= − dv = . (94) 4 r a 2 4πϵ b 8πϵ 2 0 0 a a If b = a then U = 0, of course, because the shells are right on top of each other, so the charges cancel and we effectively have no charge anywhere in the system. If b → ∞ then U = Q2 /8πϵ0 a, which is correctly the energy of a single shell with charge Q (see Problem 1.32). If a → 0 then U correctly goes to infinity, because the field diverges (sufficiently quickly) near a point charge. Equivalently, it takes an infinite amount of energy to compress a given amount of charge down to a point. The result in Eq. (94) can be interpreted as follows. As mentioned above, the energy stored in a system consisting of one spherical shell of radius r is Q2 /8πϵ0 r. Given this result, consider building up the present two-sphere system from scratch (that is, by bringing charges in from infinity) in two steps. It takes an energy Q2 /8πϵ0 a to construct the shell of radius a. Then, with that shell in place, it takes an energy Q2 /8πϵ0 b − Q2 /4πϵ0 b to construct the outer shell of radius b. The first term comes from the self energy of this outer shell. The second term comes from the potential energy of the negative outer shell due to the positive inner shell already in place (which acts like a point charge at its center). The sum of the energies of these two steps yields the result in Eq. (94). (b) Now let’s imagine starting with two neutral shells and then gradually transferring positive charge from the outer shell to the inner shell. At the start, there is no electric field between the shells, so it takes no work to transfer an initial bit of charge dq. But as more charge is piled onto the inner shell, the field grows, and it takes more work to bring in the successive bits dq. At a moment when there is charge q on the inner shell, the field between the shells is q/4πϵ0 r2 , so the force on a little charge dq is q dq/4πϵ0 r2 . The work you must do on this dq is the integral of your force times the displacement, or ) ( ∫ a 1 q dq 1 −q dq dW = − dr = , (95) 2 4πϵ0 a b b 4πϵ0 r where we have included the minus sign in the force because your force points inward (it is opposite to the electric force). However, you can always put the sign in by hand at the end; you certainly have to do positive work to move the positive charge dq toward the positively charged inner shell. We must now integrate the above work dW over all the bits dq that we bring in. This gives ) ( ( ∫ Q ) 1 1 1 q dq 1 Q2 W = − − , (96) = a b b 8πϵ0 a 0 4πϵ0 in agreement with the result in part (a). It may seem mysterious that the potential energy of a system can be found by integrating ϵ0 E 2 /2 over the volume. But the agreement of the two above methods, applied to our setup involving two shells, should help convince you that the ϵ0 E 2 /2 method does indeed give the energy. Potential energy of a cylinder The electric field at radius r outside the given cylinder equals λ/2πϵ0 r, where λ = πa2 ρ. Consider a length ℓ of the cylinder. The energy stored in the external field within this length, out to a radius R, equals Uext = ϵ0 2 ∫ R a ( πa2 ρ 2πϵ0 r )2 (2πr dr)ℓ = πρ2 a4 ℓ 4ϵ0 ∫ R a ( ) dr πρ2 a4 ℓ R . = ln a r 4ϵ0 (97) The field at radius r inside the cylinder is due only to the charge inside radius r. This charge has linear density πr2 ρ, so the field equals (πr2 ρ)/2πϵ0 r = ρr/2ϵ0 . The energy stored in the internal field, within a length ℓ, is then Uint = ϵ0 2 ∫ 0 a( ρr 2ϵ0 )2 (2πr dr)ℓ = πρ2 ℓ 4ϵ0 ∫ a 0 r3 dr = πρ2 a4 ℓ . 16ϵ0 (98) ρ = λ/πa)2 , Finding the energy per unit length simply involves erasing the ℓ. Using ( we can write the sum of Uext and Uint , per unit length, as (λ2 /4πϵ0 ) ln(R/a) + 1/4 , as desired. As mentioned in the statement of the exercise, this diverges as R → ∞. It also diverges as a → 0.