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Introduction to Electrodynamics - ch03

# Introduction to Electrodynamics - ch03 - Chapter 3 Special...

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Chapter 3 Special Techniques Problem 3.1 The argument is exactly the same as in Sect. 3.1.4, except that since z < R, vz2 + R2 - 2zR = (R - z), . q 1 ~ q. Insteadof (z - R). HenceVave= _ 4 2 R [(z + R) - (R - z)] = _ 4 R ' If there IS more than one charge 7r€o z 7r€o inside the sphere, the average potential due to interior charges is _ 4 1 Q R enc, and the average due to exterior 7r€o charges is Vcentenso Vave= Vcenter + b. .( Problem 3.2 A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV. But we know that Laplace's equation allows no local minima for V. What looks like a minimum, in the figure, must in fact be a saddle point, and the box "leaks" through the center of each face. Problem 3.3 Laplace's equation in spherical coordinates, for V dependent only on r, reads: I I 1 I 1 d ( dV ) dV dV c I c I '\72V= r2dr r2 dr =0~r2 dr =c(constant) ~ dr = r2 ~ V=-;+k. Example: potential of a uniformly charged sphere. . . 2 1 d ( dV ) dV dV c I I In cylIndrical coordInates: '\7 V = -; ds ad; = 0 ~ s ds = c ~ d; = -; ~ V = c In s + k. Example: potential of a long wire. Problem 3.4 Same as proof of second uniqueness theorem, up to the equation is V3E3 . da = - Jv(E3)2 dr. But on each surface, either V3 = 0 (if V is specifiedon the surface), or else E3.l. = 0 (if ~~ = -El. is specified). SO JV(E3)2 = 0, and henceE2 = El' qed Problem 3.5 Putting U = T = V3 into Green's identity: r [V3'\72V3 + VV3 . VV3] dr = 1 V3 vv3. da. But '\72V3 = '\72V1 - '\72V2 = -P- + P-= 0,andVV3= -E3- lv Js €o €o . So Iv E~dr = - Is V2E3 . da, and the rest is the same as before. 42

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43 Problem 3.6 Place image charges +2q at z = -d and -q at z = -3d. Total forceon +q is q [ -2q 2q -q ] ~ q2 ( 1 1 1 ) ~ I 1 ( 29q2 ) ~ F = 411"/;0 (2d)2 + (4d)2 + (6d)2 z = 47r€0d2 -2 + 8 - 36 Z = - 47r€0 72d2 Z. Problem 3.7 (a) From Fig. 3.13: -'l- = Vr2 + a2 - 2racos8j -'l-' = vr2 + b2 --2rbcos-jJ-: Therefore: , R q W -; = - - (Eq.3.15),while b = - (Eq.3.16). -'l- a Vr2 + b2 - 2rbcos8 a q q = - (Ii) vr2 + ~ - 2r~2 cos8 = - V([;.)2 + R2 - 2racos8' Therefore: 1 ( q q' ) q { 1 1 } V(r 8) - - - + - = - - , - 47r€0 -'l- -'l-' 47r€0 vr2 + a2 - 2ra cas 8 V R2 + (ral R)2 - 2ra cas 8 . Clearly,whenr = R, V ~ O. (b) u = -€o ~~ (Eq. 2.49). In this case, ~~ = ~~ at the point r = R. Therefore, u(O) = -€o (4;€0) { -~(r2 + a2 - 2racos8)-3/2(2r - 2acos8) + ~ (R2 + (ralR)2 - 2ra cos8)-3/2 (~: 2r - 2a cos8) } Ir=R = - 4~ {-(R2 + a2 - 2Racos8)-3/2(R - a cos8) + (R2 + a2 - 2Ra COS 8)-3/2 (~ - a cos 8) } = !;(R2 + a2- 2Racos8)-3/2[R - acos8 - ~ + acos8] . = I ~(R2 - a2)(R2 + a2 - 2Racos8)-3/2 .1 Qinduced= ! uda = 4;R(R2 - a2) !(R2 + a2 - 2Racos8)-3/2 R2 sinO dOdcP = -.!L(R2 - a2)27rR2 [ -2..(R2 + a2 - 2RaCOS8)-1/2 ] 1 1r 47rR Ra 0 - .!L(a2-R2) [ 1 - 1 ] - 2a VR2 + a2 + 2Ra VR2 + a2 - 2Ra . But a > R (else q wouldbe inside), so VR2 + a2 - 2Ra = a - R. = 2~ (a2 - R2) [(a ~ R) - (a ~ R)] = 2~ [(a - R) - (a + R)] = 2~ (-2R) = l-q:=ql.1
44 CHAPTER 3, SPECIAL TECHNIQUES (c) The force on q, due to the sphere, is the same as the force of the image charge q', to wit: F = ~ qq' = ~ ( - R q2 ) 1 = - ~ q2Ra 47rEO(a-b)2 47rEO a (a-R2Ja)2 47r€0(a2-R2)2' To bring q in from infinity to a, then, we do work q2R J a 0; q2 R [ 1 1 ] l a I 1 q2R W = 47r€0 (0;2 - R2)2 aa = 47r€0 - 2 (0;2 - R2) 00 = - 47r€0 2(a2 - R2)' 00 Problem 3.8 Place a second image charge, q", at the center of the sphere; this will not alter the fact that the sphere is an equipotential, 1 q" from zero to Vo = - R j 47r€0 ." q a-b ~ 8, 8 q q 'V ' a but merely increase that potential I q" = 47r€0 VoR at center of sphere.!

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Introduction to Electrodynamics - ch03 - Chapter 3 Special...

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