文档介绍：Chapter 3
Special Techniques
Problem
The argument is exactly the same as in Sect. , except that since z < R, vz2 + R2 - 2zR = (R - z),
. q 1 q.
Insteadof (z - R). HenceVave= _ [(z+ R)- (R- z)]= _ ' If there IS more than one charge
47r€o2z R ~47r€oR
1 Q
inside the sphere, the average potential due to interior charges is enc, and the average due to exterior
_47r€o R
charges is Vcentenso Vave= Vcenter+ b. .(
I
Problem I
A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV. 1
But we know that Laplace's equation allows no local minima for V. What looks like a minimum, in the figure, I
must in fact be a saddle point, and the box "leaks" through the center of each face.
Problem
Laplace's equation in spherical coordinates, for V dependent only on r, reads:
1 d dV
'\72V= r2dr r2 dr =0~r2 dr =c(constant) ~ dr = r2 ~ V=-;+k.
( ) dV dV c I c I
Example: potential of a uniformly charged sphere.
. . 2 1 d dV
dV dV c I I
In cylIndrical coordInates: '\7 V = -;ds (ad; ) = 0 ~ s ds = c ~ d; = -; ~ V = c In s + k.
Example: potential of a long wire.
Problem
Same as proof of second uniqueness theorem, up to the equation is V3E3 . da = - Jv(E3)2 dr. But on
each surface, either V3 = 0 (if V is specifiedon the surface), or else .= 0 (if ~~ = -El. is specified). SO
JV(E3)2 = 0, and henceE2 = El' qed
Problem
Putting U = T = V3 into Green's identity:
[V3'\72V3+ VV3 . VV3] dr V3vv3. da. But '\72V3 '\72V1 - '\72V2= -P- + P-= 0,andVV3= -E3-
lvr = Js1 = €o €o .
So Iv E~dr = - Is V2E3. da, and the rest is the same as before.
42
43
Problem
Place image charges +2q at z = -d and -q at z = -3d. Total forceon +q is
1 1 1 ~
q -2q 2q -q ~ q2 1 29q2 ~
F = 411"/;0[ (2d)2 + (4d)2 + (6d)2] z = 47r€0d2 (-2 + 8 - 36) Z = I - 47r€0 (72d2 )Z.
Problem
(a) From Fig. : -'l-= Vr2 + a2 - 2racos8j -'l-'= vr2 + b2 --2rbcos-jJ-: Therefore:
R q W
-;, = - - (Eq