文档介绍：Chapter 10
Potentials and Fields
Problem
2 oL 2 02V 0 02V 2 0 1
0 V + 7ft = 'VV - + ot(V .A)+ = 'VV + ot(V .A) = - fO p. ,(
2 2 02 A OV
0 A-VL = 'VA- Ot2 -V (+°{jt ) =-.,(
Problem
(a) W = ~ / (foE2 + :0 B2) dr. At tl =dJc, x;:::d = ctl, so E = 0, B = 0, and henceI W(td =
At T2 = (d + h)Jc, ct2 = d + h:
E = -~O:(d+h-x)z, B = ~~O:(d+h-x)y,
1
so B2 - -E2 and
- c2 '
fOE2 + ~B2 = fO E2 + ~.!..E2 = 2€oE2.
( ) ( €0C2 )
Therefore
(d+h)
2 2 d+h
1 :2 _(d+h-x)3 _ ~o:2lwh3
W(t2) = -(2€0)--L- (d+h-x)2dx(lw) = :lw
2 4 4 [ 3 ] - I
/d d 12
(b) S(x) = ~(B x E) ~E2[-z x (:f:y)]= :f:~E2 x = :f::2(ct - Ixl)2X
= 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
180 CHAPTER 10. POTENTIALS AND FIELDS
t2 (d+h)le
(ct - d)3 (d+h)le =
(c) W = f Pdt = /LO~:lW f (ct - d)2 dt = /LO~:lW [ ]
it dlx 3c die
Since l/c2 = /LoEo, this agrees with the answer to (a).
Problem
E=-VV_aA at= B=VXA=***@J
This is a funny set of potentials for a I stationary point charge I q at the origin. (V = 41TEO~ r~, A = 0 would,of
course, be the customary choice.) Evidently I p = qJ3(r); J =
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)
Hence = 0,,(, = 0..(.
VxE = z :x [Aowcos(kx - wi)] = -Aowksin(kx - wt) Z, - ~~ = - Aowk sin(kx - wt) Z,
aB
so VxE = -- ..(.
at
VxB = -y :x [Aokcos(kx- wi)]= Aok2sin(kx- wt)y, ~~ = Aow2sin(kx - wt) y.
So V x B = /LoEo ~ provided I k2 = /LoEOW2,lor, since C2 = 1/ /LoEo,I w =
Problem
1 1 ~
I a>.. 1 q q I 1 qt ~ fI)l
V =V--=O- --- = --; A =A+V>"=---r+ --qt --r = .LQ:J
at ( 41TEo r ) ~41TEOr 41TEOr2 ( 47r