文档介绍：Chapter 7
Electrodynamics
Problem
(a)Let Q be the charge on the inner shell. Then E = 4;EO~f in the spacebetweenthem, and (Va- Vb)=
- faE. dr - L fa J dr - ...sL 1 - 1
Jb - 41rEO Q Jb r~ - 41rEO (a b' )
1= J .da = u E. da =u Q = u 47r€o(Va- v,,) = 47rU (Va - Vb) .
I I 100 100 (I/a - I/b) 1 (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. essiveshells, as you go out, contribute less and less, because the
2 = 1
cross-sectionalarea (47rr2)gets larger and larger. For the two submer ged spheres, R = _41rua 2_1rua (one R as
the current leaves the first, one R as it converges on the second). Therefore I = V/ R = I
Problem
(a) V = Q/C = IR. Because positive I means the charge on the capacitor is decreasing,
~~= -I = - RICQ,so Q(t) = Qoe-t/ Qo= Q(D)= avo, so I Q(t) = CVoe-t/
dQ 1 o
Hence let) = --'- = CVi -e-t/RC = -e-t/RC
dt 0 RC ~R'
roo roo ~2 roo
(b)W=I! 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/
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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=
Since I(t) is the same as in (a), the energy delivered to the resistor is again I ~ 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
(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