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Introduction to Electrodynamics - ch10

# Introduction to Electrodynamics - ch10 - Chapter 10...

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Chapter 10 Potentials and Fields Problem 10.1 2 oL 2 02V 0 02V 2 0 1 0 V + 7ft = 'V V - J.LOfO Ot2 + ot (V .A)+ J.LOfo Ot2 = 'V V + ot (V .A) = - fO p. ,( 2 2 02 A ( OV ) 0 A-VL = 'V A-J.LOfO Ot2 -V V.A+J.LOf°{jt =-J.LOJ.,( Problem 10.2 (a) W = ~ / (foE2 + :0 B2) dr. At tl = dJc, x;::: d = ctl, so E = 0, B = 0, and hence I W(td = 0.1 At T2 = (d + h)Jc, ct2 = d + h: E = -J.L~O:(d+h-x)z, B = ~J.L~O:(d+h-x)y, 1 so B2 - -E2 and - c2 ' ( fOE2 + ~B2 ) = fO ( E2 + ~.!..E2 ) = 2€oE2. J.L0 J.L0€0 C2 Therefore (d+h) 1 J.L2o:2 / 2 2 d+h W(t2) = -(2€0)--L- (d+h-x)2dx(lw) = foJ.LOO: lw [ _(d+h-x)3 ] _ I foJ.L~o:2lwh3 2 4 4 3 - d d 12 (b) S(x) = ~(B x E) = ~E2[-z x (:f:y)] = :f:~E2 x = :f:J.LOo:2 (ct - Ixl)2X J.L0 J.LoC J.LoC 4c (plus sign for x > 0, as here). For Ixl > ct, S = o. So the energy per unit time entering the box in this time interval is dW = p = / S(d) . da = I f!:.oo:2lw (ct - d)2. dt L 4c Note that no energy flows out the top, since S(d + h) = o. 179

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180 CHAPTER 10. POTENTIALS AND FIELDS t2 (d+h)le (c) W = f Pdt = /LO~:lW f (ct - d)2 dt = /LO~:lW [ (ct - d)3 ] (d+h)le = it dlx 3c die Since l/c2 = /LoEo, this agrees with the answer to (a). Problem 10.3 E=-VV_aA at= [email protected] This is a funny set of potentials for a I stationary point charge I q at the origin. (V = ~ ~, A = 0 would, of 41TEO r course, be the customary choice.) Evidently I p = qJ3(r); J = 0.1 Problem lOA E = - VV - ~~ = -Ao cos(kx - wt) y( -w) = I Aowcos(kx - wt) y, I B = VXA=Z:X [Aosin(kx-wt)] =IAokcos(kx-wt)z.1 Hence V.E = 0,,(, V.B = 0..(. VxE = z :x [Aowcos(kx - wi)] = -Aowksin(kx - wt) Z, aB so VxE = -- ..(. at - ~~ = - Aowk sin(kx - wt) Z, VxB = -y :x [Aokcos(kx - wi)] = Aok2 sin(kx- wt)y, ~~ = Aow2 sin(kx - wt) y. So V x B = /LoEo ~ provided I k2 = /LoEOW2, lor, since C2 = 1/ /LoEo, I w = ck.1 Problem 10.5 I a>.. ( 1 q ) ~ q I 1 qt ~ ( 1 )( 1 ~ ) fI)l V =V--=O- --- = --; A =A+V>"=---r+ --qt --r = . LQ:J at 41TEo r 41TEOr 41TEOr2 47r£0 r2 This gauge function transforms the "funny" potentials of Frob. 10.3 into the "ordinary" potentials of a sta- tionary point charge. Problem 10.6 Ex. 10.1: V.A = 0; ~~ = o. I Both Coulomb and Lorentz.! qt ( r ) qt 3 aV I' I Frob. 10.3: V.A = -- V. 2" = --15 (r); ~ = O. NeIther. 41TEO r EO vt aV ~ Frob. 10.4:V.A = 0; at = o. ~
181 Problem 10.7 8V 8V . Suppose V.A i: -Jl.OE°Ft, (Let V.A + Jl.QE°Ft = cI>-some known functiOn.) 8V' that A' and V' (Eq. 10.7) do obey V.A' = -Jl.oEo 8t . I . W 2 W ~A 2 V.A +j.toEO 8t =V.A+V A+Jl.QE°Ft-Jl.QEO8t2 =cI>+0 A, We want to pick A such This will be zero provided we pick for Athe solution to 02 A = -cI>,which by hypothesis (and in fact) weknow how to solve, We could always find a gauge in which V' = 0, simply by picking A= J; V dt'. We cannot in general pick A = O-this would make B = 0. [Findingsuch a gauge function would amount to expressing A as - V A,and we know that vector functions cannot in general be written as gradients-only if they happen to have curl zero, which A (ordinarily) does not.] Problem 10.8 l.From the product rule: V. (~) = ~ (V .J) + J ' ( V ~) , V'. (~) = ~(V/.J) +J, (V/~). ButV~ = -V/~, since4=r-r/, So 1- 1- V. (~) = ~ (V .J) - J . (VI ~) = ~(V .J) + ~ (V' .J) - V'. (~) , But V.J = 8J% + 8Jy + 8Jz = 8J%8tr + 8Jy 8tr + 8Jz 8tr, 8x 8y 8z 8tr 8x 8tr 8y 8tr 8z and 8tr 1 & -.= ---, 8x c8x 8tr - 1 81- - - ---, 8y c8y 8tr - 1 81- 8z - -~8z' so V.J = -~ [ 8J% & + 8Jy & + 8Jz & ] = -~ 8J . (V1-). c 8tr 8x 8tr 8y 8tr 8z c 8tr Similarly, V/.J = - 8p - ~ 8J . (V/1-). 8t c 8tr [Thefirst term arises when we differentiate with respect to the explicit r/, and use the continuity equation.] thus V.

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