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Introduction to Electrodynamics - ch07 - Chapter 7...

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Chapter 7 Electrodynamics Problem 7.1 (a)Let Q be the charge on the inner shell. Then E = 4;EO ~f in the spacebetweenthem, and (Va - Vb) = - faE. dr - L Q fa J dr - ...sL ( 1 - 1 ) Jb - 41rEO Jb r~ - 41rEO a b' 1= I J . da = u I E. da = u Q = u 47r€o(Va - v,,) = 1 47rU (Va - Vb) . 100 100 (I/a - I/b) (I/a - I/b) (b) R = Va - v" = I ~ ( ! - ! ) . I 47ru a b (c)Forlarge b (b » a), the second term is negligible,and R = I/47rua. Essentially all of the resistance is in theregionright around the inner sphere. Successiveshells,as you go out, contribute less and less, because the cross-sectional area (47rr2)gets lar ger and larger. For the two submer g ed spheres, R = _ 4 2 = _ 2 1 ( one R as 1rua 1rua the current leaves the first, one R as it converges on the second). Therefore I = V/ R = I 27ruaV.I Problem 7.2 (a) V = Q/C = IR. Because positive I means the charge on the capacitor is decreasing, ~~= -I = - RICQ, so Q(t) = Qoe-t/RC. But Qo= Q(D) = avo, so I Q(t) = CVoe-t/Rc.1 dQ 1 ~ o Hence let) = --'- = CVi -e-t/RC = -e-t/RC dt 0 RC R' roo roo ~2 roo (b)W=I!cVl.1 The energydeliveredto the resistoris Jo Pdt= Jo 12Rdt= ~ Jo e-2t/RCdt= ~ (- ~C e-2t/RC) [ = ~CV02. ./ (c) Vo = Q/C + IR. This time positive I means Q is increasing: ~~ = I = RIC(CVo - Q) => Q ~~Vo = 1 1 -liCdt => In(Q - aVo) = - RCt + constant => Q(t) = avo + ke-t/RC. But Q(O) = 0 => k = -avo, so IQ(t) = avo (1 - e-t/RC) .1 let) = ~~ = avo (Rlce-t/RC) = I ~ e-t/Rc.1 125
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126 CHAPTER 7. ELECTRODYNAMICS rOO V,2 roo V,2 00 V,2 (d) Energy from battery: Jo VoIdt= ~ Jo e-t/RCdt= ~ (-RCe-t/RC)lo = ~RC=ICV02.1 Since I(t) is the same as in (a), the energy delivered to the resistor is again I ~CV02 .1 The final energyin the capacitor is also I ~ cvl, I so I half I the energy from the battery goes to the capacitor, and the other half to the resistor. Problem 7.3 (a) I = IJ . da, where the integral is taken over a.surface enclosing the positively charged conductor. But J = O'E, and Gauss's law says IE. da = (10 Q, so I = a IE. da = tiQ. But Q = CV, and V = IR, so 6£] 0 1= !LCIR, or R = _ C ' qed (0 a (b) Q = CV = CIR => ~ = -I = - JcQ => I Q(t) = Qoe-t/RC 1 ' or, since V = Q/C, V(t) = Voe-t/RC. The time constant is T = RC = ~ Problem 7.4 1= J(s) 27rsL => J(s)=I/27rsL. E=J/O'=I/27rsO'L=I/27rkL. V = -la E.dl= -~(a-b). SOIR= ~.I Problem 7.5 I=~; P=I2R= £2R . dP _£2 [ 1 2R ] r + R (r + R)2' dR - (r + R)2 - (r + R)3 = 0 =>r + R = 2R => I R =r.1 Problem 7.6 £ = f E . dl = Izero Ifor all electrostatic fields. It looks as though [; = f E . dl = (a /£o)h, as would indeed be the case if the field were really just a/Eo inside and zero outside. But in fact there is always a "fringing field" at the edges (Fig. 4.31), and this is evidently just right to kill off the contribution from the left end of the loop. The current is ~ Problem 7.7 (a) £ = -~r = -BZ~~ = -Blv; £ = IR => II = B~v.1 (Never mind the minus sign-it just tells you the direction of flow: (v X B) is upward, in the bar, so downward through the resistor.) I B2Z2V I (b)F=IlB=~, to the I left. I dv B2Z2 dv B2Z2 I 82,2 I (c) F = ma = m- = --v =>- = -( -)v => v = voe--;;;]'f"t. dt R dt Rm (d) The energy goes into heat in the resistor. The power deliveredto resistor is 12R, so dW B2Z2V2 B2Z2 B2Z2 - =I2R= -R= -v2e-2at wherea=-. dt R2 R 0' - mR' dW (it = amv5e-2at.
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