GRIFFITHS ELECTRODYNAMICS SOLUTIONS KALMAN KNIZHNIKProblem 2.1 a) Twelve equal charges, q are situated at the corners of a regular 12-sided polygon (for instance, one on each numeral of a clock face.) What is the net force on a test charge Q at the center? The force on Q must be 0 by symmetry. The x, y, and z components of the fields due to each charge cancel, so the force vanishes. b) Suppose one of the 12 q’s is removed (the one at ”6 o’clock”). What is the force on Q? Explain your reasoning carefully. Now it is like 10 of the charges cancel, the only one that doesn’t cancel being the one directly opposite the missing charge. Thus, the force is F = 1 qQ 2 4π 0 r2 (1) pointing in the direction of the missing charge. Here, r is the distance from the center to each charge. c) Now 13 equal charges, q, are placed at the corners of a regular 13-sided polygon. What is the force on a test charge Q at the center. Again, by symmetry, the answer is still 0. d) If one of the 13 q’s is removed, what is the force on Q? Explain your reasoning. The same answer as in part b, equation 1, and for the same reason. Problem 2.2 a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges, q, a distance d apart. Check that your result is consistent with what you’d expect when z d. The force would not have components in the y or x direction, if we put the charges in the x − z plane, there is no y-component of the field at the point in question, and the x-components cancel. The z-component of the field is E= 1 2q 1 2qz ˆ sin(θ)ˆ = z z 2 2 + (d/2)2 2 + (d/2)2 )3/2 4π 0 z 4π 0 (z (2) where the factor of 2 comes from adding up the two charges. When z E= 1 2q ˆ z 4π 0 z 2 d, the electric field becomes (3) which is what you would expect, since the system acts like a point charge of charge 2q. b) Repeat part (a), only this time make the right hand charge −q instead of +q. Now the z-components cancel and the x-components don’t. Mathematically, this means replacing sin(θ) with cos(θ). E= 1 2q 1 qd ˆ cos(θ)ˆ = z x 2 4π 0 z 2 + (d/2)2 4π 0 (z 2 + (d/2)2 )3/2 (4) d, then from far where the 2 has canceled out with the d/2 that comes from the cosine. If z away it looks like a dipole, and the field reduces to that of a dipole: E= 1 qd ˆ x 4π 0 z 3 (5) Problem 2.3 Find the electric field a distance z above one end of a straight line segment of length L, which carries a uniform line charge λ. Check that your formula is consistent with what you would expect for the case z L. The electric field is L dq L dq 1 1 E= cos(θ)ˆ + x sin(θ)ˆ z (6) 2 4π 0 0 r 4π 0 0 r2 √ √ replacing dq by λdx, r2 by x2 + z 2 , cos(θ) by x/ x2 + z 2 , and sin(θ) by z/ x2 + z 2 , we get E= 1 λ 4π 0 L 0 xdx 1 ˆ x+ λ 4π 0 (x2 + z 2 )3/2 L 0 zdx ˆ z (x2 + z 2 )3/2 (7) The first integral is a u-substitution, while the second can be evaluated using your favorite computer program. 1 −1 1 L ˆ E= λ{( +√ )ˆ + √ x z} 2 (8) 2 + z2 4π 0 z L z L2 + z 2 In the limit z L, we obtain 1 λL ˆ E= z (9) 4π 0 z 2 which is what you would expect from a point charge q = λL. Problem 2.5 Find the electric field a distance z above the center of a circular loop of radius r, which carries a uniform line charge λ. E= 1 dq 4π 0 R2 (10) √ Now, dq = λdl, where dl = rdθ, while R = r2 + z 2 is the distance from the edge of the loop ˆ to the point z. Furthermore, clearly the horizontal components will cancel out, leaving only a z √ component of the electric field, meaning we need to include a factor of cos(θ) = z/ r2 + z 2 . Thus, L E= 0 λ dl cos(θ)ˆ = z 2 + z2) 4π 0 (r 2π 0 1 zrdθ ˆ z λ 2 4π 0 (r + z 2 )3/2 (11) meaning the answer is E= 2 0 (r2 λrz ˆ z 2 + z 2 )3/2 (12) Problem 2.6 Find the electric field a distance z above the center of a flat circular dis ok radius R, which carries a uniform surface charge σ. What does your formula give in the limit R → ∞? Also, check the case z R. The method is the same as in problem 2.5, except that now dq = σda = σrdrdθ. Thus, the integral is R 2π R rzdrdθ −1 1 σ zσ zσ 1 ˆ ˆ E= z= z = ( −√ )ˆ 2 z (13) 2 + z 2 )3/2 2 + z 2 )1/2 2 + z2 4π 0 0 2 0 (r 2 0 z R 0 (r 0 In the limit R → ∞, we recover the field due to an infinite conducting plane, Ez = σ/2 0 , while in the limit that z R, by Taylor expanding the square root, we recover the field between two point charges: E = Q/4π 0 z 2 , where Q = πR2 σ. a) Find the charge density ρ. Since the cube has six sides. the electric field E · da = 1 0 ρdV ⇒ E4πr2 = 3q 4π 0 R3 dV = 3q 4πr3 4π 0 R3 3 ⇒ E= qr ˆ r 4π 0 R3 (17) whereas outside the electric field is just E= Thus. Vout = r ∞ R q 4π 0 r2 ∞ r ∞ R ˆ r (18) ∞ E · dl = 1 4π 0 q 1 q ˆ · dr ˆ = r r 2 r2 4π 0 r q q qr2 1 q ˆ · drˆ = r r − + 2 3 r 8π 0 R 8π 0 R 4π 0 R (19) (20) Vin = r E · dl = r qr 1 ˆ · drˆ + r r 3 4π 0 R 4π 0 this can be rewritten as Vin = 3q qr2 q r2 − = (3 − 2 ) 2 8π 0 R 8π 0 R3 8π 0 R R (21) Taking the gradient of equations 19 and 21 trivially yields in equation 18 and 17. the charge is centered in the cube.10 A charge q sits at the back corner of a cube. one centered on the charge. For instance. centered at the origin. but this is an assumption that we cannot justify.9 Suppose the electric field in some region is found to be E = kr3 ˆ. since otherwise it is unclear what fraction of the flux is going through each face. What is the flux of E through one side? We need to surround the charge symmetrically. you might think that the flux through one face is 1/6 of the total flux. no flux goes through the three faces adjacent to the charge. (Do it in two different ways). Compute the gradient of V in each region. with the original face that we were examining being 1/4 of one face of this cube. However. and check that it yields the correct result. we can determine the flux through each face by symmetry.Problem 2.21 Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. in spherical coordir nates (k is some constant). 2 Problem 2. Method 1: 2π π R Qenc = V ρdV = 0 0 dφ 0 dθ 0 0 kR 3 5 0 kr2 r2 dr = 4π 0 kR5 2 (15) (16) Qenc = E · ds = 4πR2 = 4π 0 kR5 2 Problem 2. since they are parallel to the electric field. If we center the cube on the charge. . Inside the sphere. respectively. if make a larger cube. Now. Use infinity as your reference point. This part is simple. The flux is thus q/24 0 . the original face composes 1/24 of the cube. ρ= 0 ·E= 0 2 r 1 ∂(r2 Er ) = ∂r 0k 2 r 1 ∂(r5 ) = 5 0 kr2 ∂r (14) b) Find the total charge contained in a sphere of radius R. yielding 8 total cubes. Naively. then we are effectively adding 7 cubes of the same dimensions as the original cube. ∞ We’ll use V = 0 E · dl.41. at a. ∞ R a V = 0 E · dl = 0 (0)dr + R 1 q dr + 4π 0 r2 b ∞ (0)dr + a b 1 q dr 4π 0 r2 (28) . The third charge requires 1 q(−q) 1 −q(−q) q2 1 √ W = + = ( √ − 1) (26) 4π 0 a 4π 0 a 2 4π 0 a 2 and we have already calculated the work required for the fourth charge. once again. the work for the total configuration is: W =0+ 1 −q 2 q2 1 1 q2 1 q2 q2 + ( √ − 1) + (−2 + √ ) = √ − 2 4π 0 a 4π 0 a 2 4π 0 a 2 2 2π 0 a π 0 a (27) Problem 2. Let’s calculate the electric field for this wire: E · da = E2πsL = q 0 = λL 0 ⇒ E= λ ˆ s 2πs 0 (22) Now we integrate this over the path ds ˆ to find the potential. outer radius b. using infinity as the reference point. The second charge requires (supposing it is the negative charge) work W = −qV = −q 2 /4π 0 a. is surrounded by a thick concentric metal shell (inner radius a. s since the integral will blow up. as shown in Fig. and check that it yields the correct field. W = 1 q2 1 (−2 + √ ) 2 4π 0 a 2 (25) b) How much work does it take to assemble the whole configuration of four charges? It takes no work to bring in the first charge. Instead we integrate from s to some point a.22 Find the potential a distance s from an infinitely long straight wire that carries a uniform line charge λ. the potential is zero at infinity. Therefore. 2 b) Find the potential at the center. since. since the potential is zero at infinity. the remaining charge +q will go to the outer edge of the shell. as in Fig. At a. and at b. a) Find the surface charge density σ at R. We don’t integrate from s to ∞. the surface charge density is σR = q/A = q/4πR2 . Problem 2.Problem 2. carrying charge q. so at R.35 A metal sphere of radius R. All the charge has gone to the outer radius of the metal sphere. since it is a conductor. 2. Compute the gradient of your potential. 2. from far away and place it in the fourth corner? The potential at the corner is V = V1 + V2 + V3 = 1 −q 1 q 1 q 1 1 −q √ = + + (−2 + √ ) 4π 0 a 4π 0 a 4π 0 a 2 4π 0 a 2 (24) So then the work W = qV is. a a V = s E · dl = s λ λ a ds = ln 2 2πs 0 2π 0 s (23) Taking the gradient of -V trivially reproduces equation 22. The shell carries no net charge. causing a surface charge density σb = q/4πb2 . so the surface density is σa = −q/4πa2 . there will be induced a charge -q. How much work does it take to bring in another charge +q.48).31 a) Three charges are situated at the corners of a square (side a). In order to keep the shell neutral. where k . 2 e) Which of these answers would change if a third charge. by Gauss’s law (or due to the boundary conditions. qc . Therefore. The charge at the middle of each cavity will induce an equal and opposite charge density on the cavity surface. a) Find the surface charges σa . and σR . the remaining charge must be on the outer surface R. since we don’t have the required symmetry.49). in general. σb = −qb /4πb2 (31) Now. V = q 1 1 1 ( + − ) 2 4π 0 b R a (29) c) Now the outer surface is touched to a grounding wire. since the new charge contributes its own electric field.where the (0) corresponds to regions of E = 0. the charge on the outer surface will go to ground.18 The potential at the surface of a sphere (radius R) is given by V0 = kcos3θ. d) What is the force on qa and qb ? The electric field is due to the charges themselves. Therefore the force on each charge is 0. whichever you prefer). σb .36 Two spherical cavities. of radii a and b. and the field due to charge a does not influence charge b. outside of R: E= qa + qb ˆ r 4π 0 R2 (33) Note that this is consistent with Gauss’s Law: E · da = Qenc / 0 . the potential at the center will have no contribution from the outer part of the shell. respectively. σb = 0. At the center of each cavity a point charge is placed – call these charges qa and qb . since the sphere is overall neutral. which lowers its potential to zero (same as at infinity). Furthermore. Problem 3. since the new charge would induce some more charge on the outer surface. are hollowed out from the interior of a (neutral) conducting sphere of radius R (Fig. different from each other and the unit vector ˆ in r equation 33. σR = (qa + qb )/4πR2 (32) b) What is the field outside the conductor? ˆ The field outside a conductor is always E = σ/ 0 n. Thus. As a result. Thus. since it will go from a region of high potential to region of low potential. Therefore. c) What is the field within each cavity? Once again. and vice versa. Ea = and Eb = 1 qa ˆa ∀ r < a r 4π 0 r2 1 qb ˆb ∀ r < b r 4π 0 r2 (34) (35) Note that these unit vectors are. Note that it would be difficult to apply Gauss’s law here. it would affect ER . Let the radii of the cavities be a and b. ∞ R a V = 0 E · dl = 0 (0)dr + R 1 q dr + 4π 0 r2 b ∞ (0)dr + a b (0)dr = q 1 1 ( − ) 2 4π 0 R a (30) Problem 2. were brought near the conductor? σR would be different. How do your answers to (a) and (b) change? If the outer surface is grounded. 2. Then σa = −qa /4πa2 . B3 into the general solution. while outside the sphere. the Bl ’s must be 0. B3 = (44) 5 R R 5 5 Thus. (Assume there’s no charge inside or outside the sphere. outside. (plugging our A1 . A3 . equation 40): Vin = Vout = −3k 8k 3 rP1 (cosθ) + r P3 (cosθ) 2 5R 5R3 (45) −3kR2 1 8kR4 1 P1 (cosθ) + P3 (cosθ) 2 (46) 5 r2 5 r4 To obtain σ. θ) = l=0 Al rl Pl (cosθ) + Bl Pl (cosθ) rl+1 (40) Inside the sphere. Vin = k −3k 8k [8P3 (cosθ) − 3P1 (cosθ)] = A1 RP1 (cosθ) + A3 R3 P3 (cosθ) ⇒ A1 = . we use the fact that the discontinuity in the electric field at the boundary is equal to the surface charge density: σ 0 = −( ∂V ∂n − out ∂V ∂n )= in ∂Vin ∂r − R ∂Vout ∂R (47) R . A2 = 0. 5 y= −3 5 (38) 5cos3 θ − 3cosθ 5x 3 2y − 3x + ycos(θ) = cos θ + cosθ 2 2 2 (37) So our potential.) This potential can be decomposed into a linear combination of Legendre polynomials: V0 = kcos3θ = k[4cos3 θ − 3cosθ] = k[xP3 (cosθ) + yP1 (cosθ)] (36) I can tell immediately that it is some combination of P3 and P1 since the function is odd. is V0 = k [8P3 (cosθ) − 3P1 (cosθ)] 5 (39) Now the general solution to Laplace’s equation (in spherical coordinates) is ∞ V (r. we have. A3 = (43) 5 5R 5R3 while equating this with outside the surface yields (Bl=1. since otherwise we will have problems at infinity: ∞ ∞ Vin = l=0 Al rl Pl (cosθ). Vout = l=0 Bl Pl (cosθ) rl+1 (41) Clearly the only l’s that contribute are l = 1 and l = 3: ∞ Vin = l=0 Al rl Pl (cosθ) = k [8P3 (cosθ) − 3P1 (cosθ)] 5 ⇒ A0 = 0. So now I have the equality (if I replace P3 and P1 with the explicit forms of the polynomials): 4cos3 θ − 3cosθ = x and therefore we have 5x 2y − 3x =4 = −3 2 2 ⇒ 8 x= . at the surface of the sphere. or we will have problems at the origin.is a constant. B1 . the Al ’s must be 0. as well as the surface charge density σ(θ) on the sphere. Al = 0 ∀ l > 3 (42) and so.3 = 0): Vout = k B1 B3 −3kR2 8kR4 [8P3 (cosθ) − 3P1 (cosθ)] = 2 P1 (cosθ) + 4 P3 (cosθ) ⇒ B1 = . Find the potential inside and outside the sphere. written in terms of the Legendre polynomials. unless k = 5 to match the left hand side. Taking the derivatives of equations 52 and 53 and subtracting yields asin5φ 0 ∞ = k=1 kRk−1 (ak cos(kφ) + bk sin(kφ)) + kR−k−1 (ck cos(kφ) + dk sin(kφ)) (55) Clearly. and plugging this into equation 56 to solve for d5 asin5φ 0 = 5R4 d5 d5 sin(5φ) + 5 6 sin(5φ) R10 R ⇒ d5 = aR6 10 0 ⇒ b5 = a 10R4 (58) 0 Now we have everything we need to plug in to the potentials inside (equation 52) and outside (equation 53). we need a second equation. Use the general solution to Laplace’s equation with cylindrical symmetry: ∞ V (s. since otherwise the potential blows up at infinity. we must have b0 = ck = dk = 0. since otherwise the potential blows up at s = 0. ak = ck = 0. If we note that all the coefficients vanish except b5 and d5 : Vin = s5 a sin(5φ) 2 10R4 0 (59) .Cranking out the derivatives: ∂Vin −3k 24kr2 P3 (cosθ) = P1 (cosθ) + ∂r 5R 5R3 = R −3k 24k P1 (cosθ) + P3 (cosθ) 5R 5R = 6k 32k P1 (cosθ) − P3 (cosθ) 5R 5R (48) ∂Vout 32kR4 6kR2 P1 (cosθ) − P3 (cosθ) = ∂r 5r3 5r5 Subtracting one from the other yields: σ= 0 (49) R −9k P1 (cosθ) + 5R 0 56k P3 (cosθ) 2 5R (50) Problem 3. The second equation is the continuity of the potential inside and outside the cylinder Vin |R = Vout |R : a0 + R5 b5 sin(5φ) = A0 + R−5 d5 sin(5φ) ⇒ a0 = A0 . b5 = d5 R10 (57) We can choose a0 = A0 = 0. ∞ Vin (s. φ) = A0 + k=1 s−k (ck cos(kφ) + dk sin(kφ)) (53) Using the definition σ 0 = −( ∂V ∂n − out ∂V ∂n )= in ∂Vin ∂Vout − ∂s ∂s (54) s=R where σ is given by σ(φ) = asin5φ. φ) = a0 + b0 ln(s) + k=1 [sk (ak cos(kφ) + bk sin(kφ)) + s−k (ck cos(kφ) + dk sin(kφ)] (51) Inside the cylinder. asin5φ 0 = 5R4 b5 sin(5φ) + 5R−6 d5 sin(5φ) (56) In order to calculate b5 and d5 .25 Charge density σ(φ) = asin5φ (where a is a constant) is glued over the surface of an infinite cylinder of radius R. Outside we require b0 = ak = bk = 0. φ) = a0 + k=1 ∞ sk (ak cos(kφ) + bk sin(kφ)) (52) Vout (s. So. Find the potential inside and outside the cylinder. the potential at a point on the z-axis is 1 kπ 2 R5 V (z) = 2 (67) 4π 0 48z 3 where I have let the original r → z.27 Four particles are placed as follows: charge q at (0. so the dipole term is also zero. (the monopole term) the potential vanishes. so the dipole moment is p = (3q)(a)ˆ + (q)(−a)ˆ + (−2q)(−a)ˆ + (−2q)(a)ˆ + 0ˆ = 2qaˆ z z y y x z So the dipole potential is Vdip = 1 2qaˆ · ˆ z r 1 2qacosθ = 2 2 4π 0 r 4π 0 r2 (70) (69) 1 1 4π 0 r3 3cos2 θ − 1 2πkR ρdV = 2 8π 0 r3 π R . far from the sphere. 0. a). The dipole term is Vdip (r) = 1 1 4π 0 r2 rcos(θ)ρdV = 1 2π 4π 0 r2 π 0 0 R rcos(θ)k R (R − 2r)sin(θ)r2 sin(θ)drdθ r2 (65) and we can see immediately that the θ integral vanishes. Vmon = 0. To first order. since the total charge is zero. The quadruple term won’t be zero: 1 (R − 2r)sin(θ)r2 drdθ r2 0 0 (66) This is not a difficult integral. charges −2q at (0. a. carries charge density ρ(r. θ are the usual spherical coordinates. Express your answer in spherical coordinates. and charge 3q at (0. centered at the origin.Vout = s−5 aR6 sin(5φ) 2 10 0 (60) Problem 3. Find the approximate potential for points on the z axis. Vquad = r2 r2 (3cos2 θ − 1) Problem 3. −a). 0). and it equals kπ 2 R5 /48. −a. Vdip = 0. Let’s examine the dipole term: 1 p·ˆ r Vdip = (68) 4π 0 r2 But p = qi di . 0.26 A sphere of radius R. For reference. θ) = k R (R − 2r)sin(θ) r2 (61) where k is a constant. so there is no monopole term. the exact form of the multipole expansion potential is V (r) = 1 4π 0 ∞ n=0 1 rn+1 rn Pn (cosθ)ρ(r)dV (62) The monopole term in the potential is Vmon (r) = We need to find Q. π R 1 Q 4π 0 r (63) Q= ρdV = 2π 0 0 k R (R − 2r)sin(θ)r2 sin(θ)drdθ = kπ 2 r2 R 0 (R2 − 2r)dr = 0 (64) Thus the total charge on the sphere is 0. 0) and (0. Find a simple approximate formula for the potential valid at points far from the origin. Thus. and r. Q = 3q + (−q) = 2q. are separated by a distance a. 0. The monopole moment is that total charge. −q at origin. The dipole moment is p = (3q)(0) + (−q)(−a)ˆ = qaˆ 2 z z The approximate potential is V = 1 Q p·ˆ r 1 2q qacosθ ( + 2 )= ( + ) 2 4π 0 r r 4π 0 r r2 (74) (73) c) 3q at (0. which is again 2q. the dipole moment is p = (3q)(a) + (−q)(0)ˆ = 3qaˆ 2 y y meaning that the potent is V = ˆ r since y · ˆ = sinθsinφ. a)? . while the dipole moment is in the z direction. θ = 0. so pq ˆ F= z (79) 2π 0 a3 ˆ Since the sinθ kills the second term in equation 78 above. a. a). b) 3q at origin. 0. pointing in the z direction. Problem 3. 0. −q at origin. 0. −q at (0. a) 3q at (0.θ=π/2 (78) r 1 pcosθ 1 p·ˆ = 2 4π 0 r 4π 0 r2 (77) 1 Q p·ˆ r 1 2q 3qasinθsinφ ( + 2 )= ( + ) 2 4π 0 r r 4π 0 r r2 (76) (75) ˆ since for θ = π/2. 0. For each of the arrangements below. 0) (Cartesian coordinates)? The potential on the x-axis due to a dipole is V = So the force is F = −q V = −q ∂V q ∂V ˆ q 2pcosθ 1 qpsinθ ˆ pq ˆ+ ˆ ˆ− r θ= r θ =− z 3 3 ∂r r ∂θ 4π 0 r 4π 0 r 4π 0 a3 r=a.30 Two point charges. b) What is the force on q at (0.31 A ”pure” dipole p is situated at the origin. (ii) the dipole moment. a)? This time. 0. the unit vector θ is in the -z direction. By the way. a) What is the force on a point charge q at (a. The monopole moment is just the charge. 0) to (0. and (iii) the approximate potential (in spherical coordinates) at large r (include both the monopole and dipole contributions). find (i) the monopole moment. The monopole moment is the total charge 2q. r c) How much work does it take to move q from (a. θ = π/2 since the point we are looking at is on the x axis. 3q and −q. The dipole moment is p = qi di p = (−q)(0) + (3q)(a)ˆ = 3qaˆ 2 z z The approximate potential is the sum of the monopole and dipole terms: V = 1 Q p·ˆ r 1 2q 3qacosθ ( + 2 )= ( + ) 2 4π 0 r r 4π 0 r r2 (72) (71) ˆ r where I have replaced z · ˆ with cosθ. and the ˆ direction is in z.Problem 3. 0). −a). monopole and dipole terms included. the dipole potential is V = 1 4π 0 qaˆ · ˆ z r 1 qa = cosθ 2 r 4π 0 r2 (84) Thus the total potential. is E=− V = q −1 a ˆ [ 2 ˆ + 3 (2cosθˆ + sinθθ)] 2 r r 4π 0 r r (86) 1 qa 1 q + cosθ 2 4π 0 r 4π 0 r2 (85) Problem 4. Express your answer in spherical coordinates. pq (80) W = 4π 0 a2 Problem 3. inside the sphere. Find the approximate electric field at points far from the origin. is: V =− The electric field.38. We need to find the electric field everywhere first. Find the potential at the center of the sphere (relative to infinity). The first order term in the potential is V = −q/4π 0 r2 . each a distance a from the origin. the electric field is Eout = D = D 0 4 ρdV = ρ πr3 3 ⇒ 1 Din = ρrˆ r 3 (87) 4 ρdV = ρ πR3 3 D D 0 r ⇒ Dout = 1 ρR3 ˆ r 3r2 (88) = = ρr 3 0 r ˆ r (89) = ρR3 ˆ r 3 0 r2 (90) .20 A sphere of linear dielectric material has embedded in it a uniform free charge density ρ. if its radius is R and its dielectric constant is r . Thus. therefore. D · da = D4πr2 = Qf = while outside the sphere.This is just the difference between the answers to parts a and b. since W = qV .32 Three point charges are located as shown in Fig. 3. Inside the sphere. since the net charge is -q. multiplied by q. D · da = D4πr2 = Qf = Therefore. and include the two lowest orders in the multipole expansion. The dipole term is 1 p·ˆ r 1 p·ˆ r V = = (81) 2 2 4π 0 r 4π 0 r where the dipole moment is p= ρ(r)rdV = −qδ(y + a) + qδ(z − a) − qδ(y − a)(xˆ + yˆ + zˆ)dxdydz = x y z (82) multiplying this out: p= qδ(z − a)zdzˆ − q(δ(y − a) + δ(y + a))ydyˆ = qaˆ − qaˆ − q(−a)ˆ = qaˆ z y z y y z (83) Thus. Ein = Outside the sphere. and use Eq. Find the energy of this configuration. (93) Now the polarization due to this field is P1 = E2 = − which results in the field E2 = −P1 /3 0 : (94) P1 χe χe E0 χ2 0 χe E 1 =− = − (− ) = e E0 3 0 3 0 3 3 9 Thus we obtain an infinite geometric series ∞ Etot = E0 n=0 (− E0 χe n ) = 3 1 + χe /3 (95) Using χe = r − 1.30 to write down the resulting polarization P0 . We will use equation 4.From which we can find the potential: ∞ R ∞ R V = 0 E · dl = 0 Ein · dl + R Eout · dl = 0 ρr 3 0 r ∞ dr + R ρR3 dr 3 0 r2 (91) (92) = ρR2 ρR2 ρR2 1 + = (1 + ) 2 6 0 r 3 0 3 0 2 r Problem 4. of radius a. which in turn modifies the polarization by an amount P1 .49.18: E = −P/3 0 .. E1 = − P0 χe E0 0 χe E 0 =− =− 3 0 3 0 3 0 χe E1 .23 Find the field inside a sphere of linear dielectric material in an otherwise uniform electric field E0 by the following method of successive approximations: First pretend the field inside is just E0 . Sum the series.58 in Griffiths: W = 1 2 D · Edτ (97) It is evident that for r < a. and compare your answer with Eq. The resulting field is E0 + E1 + E2 + .. Thus. 4. It is surrounded by a linear dielectric material of susceptibility χe . This polarization generates a field of its own.. and so on. The polarization due to the field E0 is P0 = 0 χe E0 . which further changes the field by an amount E2 . the energy is W = 1 2 b a Q 1 Q ˆ· ˆ4πr2 dr + r r 2 2 4πr 4π r 2 ∞ b Q Q ˆ· ˆ4πr2 dr r r 2 4πr 4π 0 r2 (101) . The field resulting from this polarization is given by equation 4. 4.26 A spherical conductor. out to radius b. carries a charge Q. E1 (Ex. we obtain: Etot = E0 = 2/3 + r /3 3 E0 2 r +2 (96) Problem 4. 4. Outside the conductor. both D and E are 0. using the fact that = 0 (1 + χe ): Q ˆ r>a D= r (98) 4πr2 Q ˆ a<r<b E= r (99) 4π r2 Q ˆ r>b E= r (100) 4π 0 r2 Thus.2). and check that they add up to zero. centered at the origin. q ˆ r 4π 0 (1 + χe )r2 χe q ˆ r 4π(1 + χe )r2 ˆ r χe q δ 3 (r) =− 2 r (1 + χe ) (108) so P= (109) The bound volume charge is ρb = − where I have used the fact that ·P=− χe q 4π(1 + χe ) · (110) · (ˆ/r2 ) = 4πδ 3 (r). the total bound surface charge is σb = 3ka. the bound surface charge on one side is ˆ σb = P · n = k(xˆ + yˆ + zˆ) · z x y z ˆ z=a/2 = ka 2 (106) Thus. Find the electric field. The electric field. and the bound charge densities ρb and σb . W = Q2 8π b a dr Q2 + 2 r 8π 0 ∞ b dr Q2 1 1 Q2 = ( + )− 2 r 8π a b 8π 0 b (102) Or.where the integral from 0 to a contributes no energy to the system.32 A point charge q is imbedded at the center of a sphere of linear dielectric material (with susceptibility χe and radius R). Let’s find the bound volume charge first: ρb = − · P = −k(1 + 1 + 1) = −3k (105) (104) Meanwhile. where k is a constant. is E= The polarization is P = 0 χe E. Thus. The polarization can be rewritten in Cartesian coordinates. writing the answer in terms of the susceptibility. Now we need to find the total charge: Qtot = ρb dV + σb da = −3k(a3 ) + 3ka(a2 ) = 0 2 (107) Problem 4. carries a ”frozen-in” polarization P = kr.31 A dielectric cube of side a. Find all the bound charges. P = k(xˆ + yˆ + zˆ) x y z We also know that − ˆ · P = ρb and P · n = σb . since there are six sides to a cube. defined such that = 0 (1 + χe ). the polarization. W = Q2 Q2 1 1 ( + )− 2 8π 0 (1 + χe ) a b 8π 0 b (103) Problem 4. What is the total bound charge on the surface? Where is the compensating negative bound charge located? The displacement field D is easily found to be D = q/4πr2 ˆ. Meanwhile the bound surface charge is r ˆ σb = P · n R = χe q 4π(1 + χe )R2 (111) Thus the total bound charge on the surface is Qsurf ace = σb da = χe q χe 4πR2 = q 2 4π(1 + χe )R 1 + χe (112) . written in terms of r the susceptibility. for r = a and r = b. r B= µ0 I π µ0 I π µ0 I 1 1 (ˆ) + z (ˆ)(− ) = z ( − )ˆ 2 z 4πa 2 4πb 2 8 a b (119) b) Here. Thus. a) If it is uniformly distributed over the surface. There is clearly no contribution from the straight sides.9 Find the magnetic field at point P for each of the steady current configurations shown in Fig. since the field flips sign. lying in the yz plane and centered at the origin. since the length perpendicular to the flow is the circumference (not the surface area. when you look down the x axis. so the total force is ˆ F = Ika2 z 2 (115) Problem 5. flowing counterclockwise. or more precisely J = α/s. what is the surface current density K? K = dI/dl⊥ . The force on both edges will conspire.5 A current I flows down a wire of radius a. a 1 ˆ dzˆ × kzˆ = Ika2 z y x F = I dl × B = I (114) 2 0 But that was only for one edge of the loop. The force on the two horizontal segments in z will cancel out. Problem 5. ˆ The force is F = Idl × B. which has a component parallel to the flow). Find the force on a square loop (side a).4 Suppose that the magnetic field in some region has the form B = kzˆ . (where k is a x constant). since here dl × ˆ = 0. so I = Kdl⊥ = K2πa.23 We will try to use the Biot Savart law: a) B= µ0 4π Idl × ˆ r µ0 I = 2 r 4π π/2 0 ˆ r adφφ × ˆ µ0 I + a2 4π 0 π/2 ˆ r bdφφ × ˆ b2 (118) ˆ where I have used that dl = rdφφ. if it carries a current I.The compensating negative charge is located at the origin (in the ρb term): Qcomp = ρb dV = − χe q (1 + χe ) δ 3 (r) = − χe q (1 + χe ) (113) which exactly cancels the surface charge in equation 112. K = I/2πa 2 b) If it is distributed in such a way that the volume current density is inversely proportional to the distance from the axis. with α some constant. a I= and therefore JdS = α 0 2πs ds = 2πaα s I 2 2πsa ⇒ α= I 2πa (116) J= (117) Problem 5. the Biot Savart integral becomes (for the semicircular side): B= µ0 I 4π π/2 −π/2 ˆ Rdφ(−φ) × ˆ r µ0 I ˆ =− z 2 R 4R (120) . what is J? We have that J ∝ 1/s. Thus. Thus. 5. The inner solenoid (radius a) has n1 turns per unit length. substituting I = λv. so the total field at s = R is: B = (− µ0 I µ0 I + )ˆ z 4R 2πR (121) Problem 5. and the wires will repel. so we will refer to Griffiths example 5.For the two straight lines. moving along at a constant speed v. (ii) between them. µ0 I1 I2 µ 0 λ2 v 2 fB = = (123) 2πd 2πd Meanwhile. the Biot Savart integral is awkward and difficult to compute. The magnitude of the force is F = ILB. the total force is infinite. where L is the length of the wire. we have contributions from both the larger and the smaller solenoid. The field in region (i) is B(i) = µ0 n1 Iˆ + µ0 n2 I(−ˆ) = µ0 I(n1 − n2 )ˆ 2 z z z (127) where the minus sign comes from the fact that the current is being carried in the opposite direction in the larger solenoid.15 Two long coaxial solenoids each carry current I. The magnetic field at a distance d due to the first (second) wire is µ0 B= I1 (I2 ) (122) 2πd There is an attractive force between the wires. the field is zero. so the force per unit length between these two wires is. Thus. Thus. but in opposite directions. as shown in Fig. 5. the field in region (ii) is B(ii) = µ0 n2 I(−ˆ) = −µ0 n2 Iˆ 2 z z (128) . How great would v have to be in order for the magnetic attraction to balance the electric repulsion? A line charge λ traveling at velocity v constitutes a current I = λv.12 Suppose you have two infinite straight line charges λ. Outside. and (iii) outside both. Problem 5. so the magnetic field due to the smaller one is zero. In region (ii). a distance d apart. 2π 0 d (124) meaning that the force per unit length between the wires is fe = λE λ2 = L 2π 0 d (125) Thus the electric force per unit length fe ≡ Fe /L can be set equal to the magnetic force per unit length: λ2 1 µ 0 λ2 v 2 = ⇒ v= =c 2 (126) fB = fe ⇒ 2πd 2π 0 d 0 µ0 Thus you could never get the wires moving fast enough – the electric force will always dominate over the magnetic force. Now. The field inside a solenoid is B = µ0 nI.42. in which he states that the magnetic field between these two wires is going to be B = µ0 I/2πs. Find B in each of the three regions: (i) inside the inner solenoid. in region (i). The region is outside the smaller solenoid. but the force per unit length is f = IB. since the current is in the same direction. the only contribution to the magnetic field is from the larger solenoid. we need to find the electric field due to a wire of charge density λ: E · dl = E2πd = λL 0 ⇒ E= λL . and the outer one (radius b) has n2 . so the force on the upper plate is F = σ 2 A/2 0 . the field adds: Bbetween = µ0 µ0 σv − (−σv) = µ0 σv (in) 2 2 2 (130) b) Find the magnetic force per unit area on the upper plate. We consider each moving plate to be a surface current K. Equating this force with the force per unit area in part (b) yields: µ0 σ 2 v 2 σ2 1 = ⇒ v=√ 2 (132) 2 2 0 µ0 0 So the plates would have to move at the speed of light.Outside both solenoids.22 Find the magnetic vector potential of a finite segment of a straight wire. a) Find the magnetic field between the plates and also above and below them.43.8. use Ampere’s law: B · dl = 2BL = µ0 Ienc = µ0 KL ⇒ B = µ0 K/2. the field is zero. c) At what speed v would the magnetic force balance the electrical force? The electric field of the bottom plate is E = σ/2 0 . The force per unit area is f = K × B. and z dl = dz. Thus. thus.64: A= µ0 4π Idl |r − r | (133) Check that your answer is consistent with equation 5. Thus. carrying a current I. where the + is below the plane and the − is above the plane. [Put the wire on the z axis. Problem 5.35. B(iii) = 0 2 (129) Problem 5.16 A large parallel-plate capacitor with uniform surface charge σ on the upper plate and −σ on the lower is moving with a constant speed v. and. The force per unit area is. the forces would never balance. Thus. in region (iii). as shown in Fig. How is this derived? Well. According to example 5. from z1 to z2 . letting I = Iˆ. the magnetic field due to such a surface current is B = ±µ0 K/2. we take the curl of A: B= ×A=− ∂Az ˆ φ ∂s (136) . f = σ 2 /2 0 . In between the capacitor plates. 2 s 2 + z 2 + z2 ˆ µ 0 I z2 zdz µ0 I √ ˆ A= = ln z 2 (135) 4π z1 4π 2 z 2 + s2 s 2 + z 1 + z1 To confirm whether this matches equation 5. 5. so we obtain. the field is zero. for theupperplate f = σv µ0 σ 2 v 2 µ0 σv = 2 2 (131) Doing the right hand rule will convince you that the force is up. where K = σv. above and below the capacitor. since the surface current K is pointed in opposite direction due to the opposite charge on the plates.35: B= µ0 I ˆ (sinθ2 − sinθ1 )φ 4πs (134) √ The distance between the wire and any point s off the wire is z 2 + s2 . and use equation 5. including its direction. The magnetic field due to the circular loop is given by everyone’s favorite formula for the magnetic field due to a dipole: µ0 Iπa2 µ0 1 ˆ [3(m · ˆ)ˆ − m] = − rr z (139) B= 4π r3 4π r3 since m · ˆ = 0. the loop is spinning in the direction of negative z. Kb = M × n = M z × ˆ = M φ (141) So the cylinder acts like a solenoid. 6. the field is also zero. on the surface. Kb = M × n = −ks2 z z s ∂s ˆ = −kR2 z s=R (142) (143) Now we can get the magnetic field inside the cylinder by using Ampere’s law.6. due to the circular loop (assume r is much larger than a or b. where n is the number of turns per unit length. since N = 1 for this cylinder. If the square loop is free to rotate.If you take this derivative. the only contribution to Ienc is from Jb : s B · dl = B2πs = µ0 Ienc = µ0 Jb da = µ0 3ksˆ · sdφˆ = 2πµ0 ks3 z z ⇒ ˆ B = µ0 ks2 φ (144) 0 . ˆ If the magnetization is parallel to the axis then we can write M = M z. so it will anti align with the other loop. Find the magnetic field (due to M) inside and outside the cylinder. Thus. Jb = ˆ ˆ ˆ s × M = 0. Then. however. The easiest way to do this is to calculate the bound currents: Jb = ×M= 1 ∂(sMφ ) ˆ ˆ ˆ z = 3ksˆ. y Problem 6. what will its equilibrium orientation be? ˆ Torque is given by N = m × B.13). ˆ N = −Ib2 x × µ0 Iπa2 I 2 a2 b2 µ0 ˆ ˆ z=− y 4π r3 4r3 (140) The final orientation of the square loop will be when there is no torque on it. Problem 6. for points inside and outside the cylinder.7 An infinitely long circular cylinder carries a uniform magnetization M parallel to its axis. But n = N/L. s is the distance from the axis. thus the product nI = N Kb . and I = Kb L. the loop will rotate until it points in -ˆ. Since the surface current is.. Since the magnetic ˆ field is in the z direction. and.8 ˆ A long circular cylinder of radius R carries a magnetization M = ks2 φ. where the magnetization is zero. you get ˆ µ0 I [ B=φ 4πs 2 But recall that sinθj = zj / zj + s2 . ˆ ˆ B = µ0 Kb z = µ0 M z = µ0 M 2 Outside the cylinder. which is m = Ib2 x. The dipole moment of the circular loop is m = Iπa2 z.. Why minus? Since the torque z is in -ˆ . since it carries a surface current in the azimuthal direction. Find the magnetic field due to M. the magnetic field is B = µ0 nI. 6. The m in the torque equation. and φ is the usual azimuthal unit vector (Fig. where k is a ˆ constant. When that happens. refers to the dipole moment of the square r ˆ loop. so z2 2 z2 + s 2 − 2 z1 2 z1 + s2 ] (137) ˆ µ0 I (sinθ2 − sinθ1 ) 2 B=φ 4πs (138) Problem 6.1 Calculate the torque exerted on the square loop shown in Fig. 16 A coaxial cable consists of two very long cylindrical tubes. of radius R. 6.18. so B = µ0 ksˆ 2 z (150) Problem 6.20) to find H. B = µ0 M. Outside. in terms of the magnetization. Find the magnetic field in the region between the tubes. so Bout = 0. 6. since the magnetization is in that direction and the surface and volume ˆ currents are in the φ directions. since B = µH = µ0 (1 + χm )H. in each case the current distributes itself uniformly over the surface (Fig. we can immediately find the magnetic field. the total current through a square Amperian loop of side length l.2. and calculate the field they produce. and the get B from Eq. H = 0. and confirm that (together. we obtain I ˆ φ (151) B = µ0 (1 + χm ) 2πs Magnetization is given by M = χm H. where M = 0. calculate the magnetization and the bound currents. ˆ Bin = µ0 ks2 φ = µ0 M 2 (145) Inside the cylinder. placed a distance R − s into the cylinder is Ienc = Jb da + Kb dl = −k(R − s)l + kR(l) = ksl (148) and therefore the magnetic field is B · dl = Bl = µ0 Ienc = µ0 ksl ⇒ Bin = µ0 ksˆ 2 z (149) ˆ It is in the z direction. locate all the bound currents. separated by linear insulating material of magnetic susceptibility chim . M = ksˆ. Inside. there is no z free current anywhere. with the free currents) they generate the correct field. Let’s calculate these: dMz ˆ ˆ ˆ Jb = − φ = −k φ. there is no current through the Amperian loop outside. (Notice that the second method is much faster. outside.or. 6.24). ˆ The bound volume current is Jb = × M and the bound surface current is Kb = M × n. Since there is no free current anywhere. since H = B/µ0 − M. 6. of course. Therefore. so M= Iχm ˆ φ 2πs (152) . Kb = M × ˆ s = kRφ (147) ds s=R Thus. and avoids any explicit reference to the bound currents. Using Ampere’s law: H · dl = If ree ⇒ H = If ree/2πs. parallel to the axis. A current I flows down the inner conductor and returns along the outer one.) Ampere’s law is H · dl = If ree. b) Use Ampere’s law (in the form of Eq. Therefore. As a check. Meanwhile. where k is constant and s is the distance from the axis. the magnetic field is zero since the total current enclosed by an Amperian loop is zero. Let’s check that: R Ienc = Jb da + Kb dl = 3ksˆ · sdφˆ − kR2 z z dl = 2πkR3 − kR2 2πR = 0 (146) 0 Problem 6. Find the magnetic field inside and outside the cylinder by two different methods: a) As in Sect.12 An infinitely long cylinder. carries a frozen-in magnetization.enc . B = 0 also. fills the entire region. or aluminum) with susceptibility χm . Hin = I s ˆ φ 2πa2 (158) Meanwhile. Thus. What is the net bound current flowing down the wire. what is the magnetic field a distance s from the axis? Find all the bound currents. B = µH = µ0 (1 + χm )H. pointing into the page. what is the current in the resistor? In . B · dl = µ0 Ienc ⇒ B2πs = µ0 I(1 + χm ) ⇒ B= µ0 I(1 + χm ) ˆ φ 2 2πs (157) Kdl = I + Iχm Iχm ˆ z · dlˆ = I + z 2πa = I(1 + χm ) 2πa 2πa (156) Problem 6. Thus. and the current is distributed uniformly. the current at that distance is Is2 /a2 . outside the wire. so Jb = while ˆ ˆ K b = M × n = χm H × n s=a I ˆ χm z πa2 = χm I (−ˆ) z 2πa (162) (163) The net bound current is Ib = Jb πa2 Kb 2πa = Iχm − Iχm = 0 2 (164) Problem 7. Thus. The free current is Jf = I/πa2 and Jb = χm Jf . Bin = I and Bout = µ0 (1 + χm )s ˆ φ 2πa2 Iµ0 ˆ φ 2πs (160) I ˆ φ 2πs (159) (161) since the susceptibility of free space vanishes. the total current is Itot = I + Thus. At a distance s < a from the center of the wire. H · dl = 2πsH = If ree . say.Therefore. a) If the bar moves to the right at speed v. A resistor R is connected across the rails and a uniform magnetic field B.7 A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance l apart. J= ˆ Ka = M × n s=a ×M= 1 ∂(sMφ ) ˆ z=0 s ∂s Iχm ˆ Iχm ˆ = = φ׈ s z 2πs 2πa s=a Iχm Iχm ˆ ˆ =− φ׈ s z 2πs 2πb s=b (153) (154) and ˆ Ka = M × n s=b = (155) Thus. Hout = Now.17 A current I flows down a long straight wire of radius a. the enclosed current is just I. Since the current is distributed uniformly. If the wire is made of linear material (copper. The bar must slow down.what direction does it flow? If the bar is moving to the right. Find the current induced in the loop. since the positive sense was defined by the direction of the magnetic field. and power is the time derivative of energy: P ≡ dW/dt: dW B 2 l2 v 2 B 2 l2 v 2 P = = I 2R = ⇒ dW = dt (170) dt R R Plugging in my result for v 2 from equation 169: W = B 2 l2 R ∞ 0 2 v0 e−2B 2 l2 t/Rm dt = 2 B 2 l2 v0 Rm 2 2 (e−2B l t/Rm ) R −2B 2 l2 ∞ 0 1 2 = mv0 2 2 (171) Problem 7. Since the current is up through the bar and the magnetic field is into the page. The area of the loop is lvt. c) If the bar starts out with speed v0 at time t = 0. 2. what is its speed at a later time t? m dv B 2 l2 v =− dt R ⇒ v(t) = v0 e−B 2 l2 t/Rm 2 (169) 2 d) The initial kinetic energy of the bar was. that means that the flux through the loop is increasing. The force is I=− F = IlB = − B 2 l2 v R (168) dφB = −Blv = IR dt (166) where the minus sign means to the left. The flux through the loop is π a φB = π( )2 B0 cos(ωt) = a2 B0 cos(ωt) 2 4 so the induced EMF is EM F = − dφB πa2 B0 ω = sin(ωt) dt 4 (172) (173) . Check the energy 2 /2. is driven by an alternating current. A circular loop of wire. mv0 /2. and is left to slide. whose magnitude is EM F = − meaning that the current is Blv 2 (167) R The negative sign means that the current is flowing counterclockwise (i. and coaxial with it. the force is to the left. so that the flux (and it’s time derivative) through the loop is φB = lvtB ⇒ dφB = Blv dt (165) By Faraday’s law. as a function of time. downward through the resistor). as it must be. or radius a/2 and resistance z R. of course. of radius a. b) What is the magnetic force on the bar? In what direction? The magnetic force is F = Idl × B. so that the field inside is sinusoidal: B(t) = B0 cos(ωt)ˆ. is placed inside the solenoid.e. delivered to the resistor is exactly mv0 The power delivered to the resistor is P = I 2 R. the bar would gain more and more energy as it sped up.12 A long solenoid. this produces an EMF in the loop. without a source of that energy. Otherwise. dI µ0 na2 dI ˆ E2πs = −µ0 nπa2 ⇒ E=− φ (180) dt 2s dt Problem 7. of width w a. and outside is 0. forms a parallel-plate capacitor. the electric field can be found through E · dl = − d ˆ µ0 nI(t)πa2 φ dt (178) B · da vanishes outside. the flux through an Amperian loop of radius s inside the solenoid is φB = µ0 nI(t)πs2 The induced EMF is EM F = −µ0 nπs2 and the electric field can be calculated via EM F = E · dl = E2πs ⇒ Ein = − µ0 ns dI 2 dt 2 (177) dI dt (175) (176) ˆ The electric field circles around the solenoid. The magnetic field inside a solenoid is B = µ0 nI.meaning that the current I = EM F/R is I= πa2 B0 ω sin(ωt) 4R (174) Problem 7.32 The preceding problem was an artificial model for the charging capacitor. radius a. Thus. Find the electric field (magnitude and direction) at a distance s from the axis (both inside and outside the solenoid). 7. the flux through an Amperian loop of radius s is φE = Thus. we have B · dl = 0 µ0 σ 0 πs2 ⇒ dφE πs2 dσ πs2 1 dq Is2 = = = 2 2 dt a 0 0 dt 0 πa dt (181) ∂φE = ∂t 0 µ0 Is a2 0 2 ⇒ B= µ0 Is2 1 ˆ µ0 Is ˆ φ= φ 2 a2 2πs 2πa2 (182) Problem 7. as shown in Fig. designed to avoid complications associated with the current spreading out over the surface of the . The displacement current is given by the change of electric flux. at a distance s < a from the axis. A narrow gap in the wire. Meanwhile. This is a displacement current problem. The electric field due to the charge density on the edge of the wire is E = σ/ 0 . Find the magnetic field in the gap. the contribution to Thus.15 A long solenoid with radius a and n turns per unit length carries a time-dependent ˆ current I(t) in the φ direction. uniformly distributed over its cross section. the integral becomes 2πsE.31 A fat wire. Thus. the flux through an Amperian loop outside the solenoid is φB = µ0 nI(t)πa2 since the magnetic field outside the solenoid is 0. (179) Since the loop of integration is taken around the point outside the solenoid. carries a constant current I. since the magnetic field is in the z direction through the solenoid.43. in the quasi static approximation. as a function of t. but this time use the cylindrical surface in Fig. The flux through this circle is this electric field multiplied by the area of the circle. We found it to be µ0 Is ˆ B= φ 2 (190) 2πa2 . 7. we have the usual electric field in a capacitor.old − Iwire = Iwire 2 − Iwire (187) a To obtain the total current to be enclosed by our Amperian loop. Using this circle as your ”Amperian loop. we again find. But q/t = I.44a). a) Find the electric and magnetic fields in the gap. the radius of the capacitor is a. as functions of the distance s from the axis and the time t. We know have a total enclosed current of (i) the current in the wire with radius s. so the electric field as a function of time is E= It ˆ z 2 π 0 a2 (183) b) Find the displacement current through a circle of radius s in the plane midway between the plates. a) Find the electric field between the plates. For a more realistic model. I: Itot = Id. πs2 : φE = Therefore. So s2 Id. Between the plates. at any given time. Again. and (ii) the displacement current from the donut region around the wire.new + Iwire = Iwire Thus. and the separation of the plates is w a. that B= µ0 Is ˆ φ 2 2πa2 (189) s2 s2 − Iwire + Iwire = Iwire 2 a2 a (188) Problem 8. which is open at the right end and extends to the left through the plate and terminates outside the capacitor. and is zero at t = 0. 7. Assume that the current flows out over the plates in such a way that the surface charge is uniform.” and the flat surface that spans it. as in equation 186. we need to add the real current flowing in the wire.44b. the current I is constant.31. the displacement current is Id = 0 It πs2 π 0 a2 (184) dφE = dt 0 Is2 πs2 = 2 2 π 0 a2 a I (185) From this we can obtain the magnetic field B · dl = µ0 Id ⇒ B2πs = µ0 Is2 a2 ⇒ B= µ0 Is ˆ φ 2 2πa2 (186) c) Repeat part (b).new = Idisp. 7. The displacement current will be the old displacement current in equation 185. find the magnetic field at a distance s from the axis.plates. and there are two contributions to Ienc .) ˙ We already found the magnetic field in problem 7. (Assume the charge is zero at t = 0. minus the current in the wire. Notice that the displacement current through this surface is zero. E = σ/ 0 = q/π 0 a2 .31 by using Maxwell’s law B · dl = 0 µ0 φE .2 Consider the charging capacitor in Prob. imagine thin wires that connect to the centers of the plates (Fig. A(r. t) = −qt/4π 0 r2 ˆ. The total power is P = S · da = I 2 wtb2 I 2t [bˆ · (2πbwˆ] = s s 2π 2 0 a4 π 0 a4 (198) Problem 10. 8. Note especially the direction of S. equation 8. the Poynting vector is trivially computed: S= Eq. c) Determine the total energy in the gap. This is a rather simple exercise: E=− V − B= ∂A 1 q ˆ = r ∂t 4π 0 r2 ×A=0 (199) (200) It looks like this corresponds to a point charge at the origin: ·E= q 4π ·( 0 ˆ r q ) = δ 3 (r) 2 r 0 ⇒ ⇒ ρ = qδ 3 (r) (201) (202) ×B=0 J=0 . so it remains to take the time derivative of equation 193 and compare it to the divergence of equation 194: µ0 I 2 ∂uem = 2 4 c2 t = − ∂t π a ·S= I 2t 2π 2 0 a4 · (sˆ) = s I 2t π 2 0 a4 (196) (194) (Recall that the divergence is taken in cylindrical coordinates).14 is satisfied. t) = r 0. 8. 8.9 – in this case W = 0. Since c2 = 1/µ0 0 . The energy density is uem = 0 2 E2 + 1 I 2 t2 1 µ0 I 2 s 2 µ0 I 2 s2 B2 = + = 2 4 (c2 t2 + ) 2µ0 2 0 π 2 a4 2 4π 2 a4 2π a 4 I 2 st 1 E × B = − 2 4ˆ s µ0 2π 0 a (193) Meanwhile. we obtain E= It ˆ z 2 2 0 πa ∂E ∂t It ˆ z 2 0 πa (191) We could have also found this by using Maxwell’s law in differential form: ×B= µ0 I 1 ∂(sBφ ) ˆ ˆ z= z= s ∂s πa2 0 µ0 ⇒ E= (192) b) Find the energy density uem and the Poynting vector S in the gap. Check that the power input is equal to the rate of increase of energy in the gap (Eq. the electric field in the gap is given by E = σ/ 0 z.ˆ Meanwhile. Using σ = q/A = It/A. Calculate the total power flowing into the gap. Check that Eq. because there is no charge in the gap). and the charge and current distributions.14 is confirmed. The total energy is given by b Uem = uem dV = 0 uem w2πsds = µ0 wI 2 b2 2 2 b2 [c t + ] 2πa4 16 (197) for an arbitrary surface of radius b.14 is ∂ (umech + uem ) = − · S (195) ∂t umech = 0 here.3 Find the fields. corresponding to V (r. as a function of time. w is the width of the gap. by integrating the Poynting vector of the appropriate surface.