p2= 由(5)(6)解得 (7) [+2mRB]s[含a+可 (8) 对(7)(8)式, 当n=0时 4nR。+a,- Q do 0 -8Q do 4π8R20 -60 Ro 解得 d。= Q 4π80 Q(8-8o) a0= 4πE8Ro 当n=1时 a,RoP= R 6a1P1= d 280 解得 d,=0 (a1=0 同理 dn=a.=0(n≥l) g=96-6 Q:1 4πeERo p2=4π6oR 55 2 n n 1 n 0 d P (cos ) R n = = + 由(5)(6)解得 f 0 Q 4 R + n n n n 0 a R P = = n n+1 n n 0 0 d P R = (7) f n n 1 2 n 2 n 0 n 0 n 0 0 n 0 n 0 Q d na R P (n 1) P 4 R R − + = = − + = − + (8) 对(7)(8)式, 当 n=0 时 f 0 2 0 0 0 f 0 2 2 0 0 0 Q d a 4 R R - Q d 4 R R + = = − 解得 f 0 0 f 0 0 0 0 Q d 4 Q ( ) a 4 R = − = 当 n=1 时 1 1 0 1 1 2 0 1 1 1 0 1 3 0 d a R P P R d a p 2 P R = = − 解得 1 1 d 0 a 0 = = 同理 n n d a 0(n 1) = = f 0 1 0 0 Q ( ) 4 R − = 2 0 1 4 R = f Q