Introduction to Electrodynamics - ch04.pdf

文档介绍：Chapter 4
Electrostatic Fields in Matter
Problem
E =V/x = 500/10-3 = 5x 105. Table : a/47r€0 = 10-30,so a = 47r( 10-12)( 10-30) =
-41. p =aE =ed ~ d =aE/e = ( 10-41)(5 x 105)/( x 10-19) = X 10-16 m.
d/R =( x 10-16)/( x 10-10) = 10- ionize, say d = R. Then R = aE/e = aV/ex ~ V =
Rex/a.= ( x 10-10)( x 10-19)(10-3)/( x 10-41) =
Problem
First find the field, at radius r, using Gauss' law: = E~Qenc,or E = 4;<0~Qenc.
r
r 47rq r - 4q a - a2
Qenc = pdT = - e-2r/ar2dr = - --e-2r/a r2 + ar +-
l0 7ra3l0 a3 [ 2 ( 2 )]l 0
2q a2 a2 r r2
= - a2 [e-2r/a (r2 + ar +"2 ) -"2 ] = q [1 - e-2r/a (1 + 2~ + 2a2)] .
[Note:Qenc(r --+ 00) =q.] So the field of the electron cloud is Ee = 4;<0~ [1 - e-2r/a(1+ 2~+ 2~)]. The
protonwill be shifted from r = 0 to the point d where Ee = E (the external field):
d ~
E=-- 1 q 1-e- 2d/ a 1+2-+2-.
47r€0d2 [ ( a a2)]
Expandingin powers of (d/a):
2d 2d 2d
e-2d/a = 1 - + ! 2 - .!. 3 +... = 1 - 2~ + 2 ~ 2 - ~ ~ 3 +...
(a ) 2 (a ) 3! (a ) a (a) 3 (a)
d d2
1- e-2d/a 1+ 2- + 2-
( a a2) = 1- (1-2~+2(~r -~(~r +..-) (1+2~+2~)
d cP. d cP. d3 cP. d3 4~
= r - r - 2ta - 2:+d2 + 2ta + 4:+d2 + 4:+d3 - 2:+d2 - 4:+d3 + -3 -a3 + .. .
4 d 3
= 3 (~) + higher order terms.
73
74 CHAPTER 4. ELECTROSTATIC FIELDS IN MATTER
1 q 4 d3
E = -- -- = -po a = 311"!:oa.
--(qd) = I I
471"€0dl- (3 a3) 471"€01 3a34 371"€oa31. 3
[Not so different from the uniform sphere model of Ex. (see Eq. ). Note that this result predicts
4;EOa = !a3 = ! ( X 10-10)3 = X 10-30 m3, compared with an experimental value (Table ) of
x 10-30 m3. Ironically the "classical" formula (Eq. ) is slightly closer to the empirical value.]
Problem
per) = Ar. Electricfield(by Gauss'sLaw): § = E (471"r2)= -!oQenc = EloJ;Ar471"r2dr, or E =
~471"r 471"€oA r44 = Ar24€0 . This "internal" field balances the external field E when nucleus is "o