由电荷分布的对称性，由高斯定理可得 是股4片照 6o 47ER3 9=∫瓦+「E,d=巴+2(c-) 4πeR'4πEsR (R<R) -医眼品 (R>R) 解法二：（分离变量法）由题意得定解问题 7201= -8L8(i-cXR<Ro) (1) 7202 0(R>R) (2) ilR-0= 有限值 (3) 02R-→o= 0 (4) 01= P2lR-Ro (5) 00- 0p2 Eo RR-Ro (6) 对(1)设9=p+0 % 则 4πER 70= 0 (1) 72p2= 0 (2) ilR-0= 有限值 (3) P2R 0 (4) 0= P2lR-Ro (5) 00三 op (6) OR OR IR=Ro 由(1)(2)解得 =2a,R+ )P,(cos) n=0 由(3)(4)解得 网=2a,RR(os0)4 由电荷分布的对称性，由高斯定理可得 1 f 2 f 1 2 3 3 0 0 D Q R D Q R , 4 R 4 R E E     = = = = ( ) ( ) 0 0 0 1 1 2 0 0 0 2 2 0 ( ) 4 4 4 R f f R R f R Q Q E dR E dR R R R R Q E dR R R R           − =  +  = +  =  =     解法二：（分离变量法）由题意得定解问题 0 0 2 f 1 0 2 2 0 1 R 0 2 R 1 2 R=R 1 2 0 R R (r c)(R R ) (1) 0(R R ) (2) (3) 0 (4) (5) (6) R R              → →  =  −  = −      =    =   =   =     =     有限值 对(1)设    1 0 1 = +  f 0 Q 4 R   = 则， 0 0 2 1 2 2 1 R 0 2 R 1 2 R=R 1 2 0 R R 0 (1) 0 (2) (3) 0 (4) (5) (6) R R           → →  =   =   =    =  =   =      =     有限值 由（1）（2）解得 n 1 n n n 1 n 0 b (a R )P (cos ) R n    =  = +  + n 2 n n n 1 n 0 d (c R )P (cos ) R n    = = +  + 由（3）（4）解得 n 1 n n n 0   a R P (cos )  =  = 