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Unformatted text preview: 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 47ro(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)For large 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 crosssectional area (47rr2)gets larger and larger. For the two submer ged 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) = Qoet/RC. But Qo= Q(D) = avo, so I Q(t) = CVoet/Rc.1 dQ 1 ~ o Hence let) = ' = CViet/RC =et/RC dt RC R' roo roo ~2 roo (b)W=I!cVl.1 The energydeliveredto the resistoris Jo Pdt= Jo 12Rdt= ~ Jo e2t/RCdt= ~ ( ~Ce2t/RC) [ = ~CV02. ./ (c) Vo = Q/C + IR. This time positive I means Q is increasing: ~~ = I = RIC(CVo Q) => Q ~~Vo = 1 1liCdt => In(Q aVo) =  RCt + constant => Q(t) = avo + ket/RC. But Q(O) = 0 => k =avo, so IQ(t) = avo (1  et/RC) .1 let) = ~~ = avo (Rlcet/RC) = I ~ et/Rc.1 125 126 CHAPTER 7. ELECTRODYNAMICS rOO V,2 roo V,2 00 V,2 (d) Energy from battery: Jo VoIdt= ~ Jo et/RCdt= ~ (RCet/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] 1= !LCIR, or R = _ C ' qed (0 a (b) Q = CV = CIR => ~ =I =  JcQ => I Q(t) = Qoet/RC 1 ' or, since V = Q/C, V(t) = Voet/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=~(ab). 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 signit 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...
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This note was uploaded on 02/17/2010 for the course PHYSICS mae142 taught by Professor Kim during the Spring '10 term at American College of Computer & Information Sciences.
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