Introduction to Electrodynamics — Instructor's Solutions by David J. Griffiths
By David J. Griffiths
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Extra resources for Introduction to Electrodynamics — Instructor's Solutions Manual
Sample text
2 coso (5COS20 - 3) = -l(l 1 1) + sin2 0 (-1O cosO sinO)] + I)P3. 17 =-1/(x)~ (5X3 - 3x) dx =~ - x3)1~1 = ~(I-I (X5 00 (a) Inside: V(r, 0) = LA,r'p'(cosO) (Eq. 66) where 1=0 1T A, = (2l + 1) 2R' / 0 . 69). In this case Vo(O) = Vo comes outside the integral, so 1T - (2l + I)Vo A, - 2R' / 0 P,(cosO) sin 0 dO. / 50 CHAPTER 3. SPECIAL TECHNIQUES But Po(cos0) = 1, so the integral can be written ". 0 0, ~f I # 0 = { 2, If I = 0 } Po(cosO)~(cosO)sinOdO (Eq. 68). Therefore ~fI # 0 . 1 The potential is constant throughout the sphere.
Let's use a sphere of radius a > R. (~ r2 sinOdOd4>+ {R E2dT + {a Jo q2 14 q2 411" 1 4 2 (411"100)2 ~ 11"+ (411"100)2 5R + (411"100)21I"q 1 1 1 1 q2 1 411"100"2~ { + 5R - ~ + R } As a --t 00, the contribution Problem 411"100 r JR (- 1 ~ 411"100 r ) 1 ( -;: ) R } l . / from the surface integral (4;€O~) 1 = dqV = dq -411"100 ~, r ( ) 4 goes to zero, while the volume integral dq 0~~ W = q R3 (q = total charge on sphere). 2 = 411"r 1 dW (q = charge on sphere of radius r). r3 q = 311"r3 p = 411"r2 dr p = 43qdr 311"R qr3 411"100R3 1 3q 2 R3 r dr ( ) ( 1 3q2 {R Jo r4dr = 411"100 R6 } a - 1)) picks up the slack.
So we must treat k = 0 separately. One solution is a constant-but what's the other? Go back to the differential equation for S, and put in k = 0: s- d ds dB ds ( ) s- dS = 0 => s-dS = constant = C => ds ds C ds = -s =>dS = C- s =>S = Clns + D (another constant). So the second solution in this case is In s. That too reduces to a single solution, = A, in the = 0 into the equation: case k = O. What's the second solution here? Well, putting k d2 d