D 于是E= 分0 EoEr 4π88，r ②∑9=fE.5=6,E4πr2= →0=29%29-9 -0 6,-10 于是 r D E  0   = 0 2 4 0 r r Q r     = ⑵   =  S q E dS   0 内  2 0 =  E 4 r r Q  = Q = q 内 − q0内 →  Q Q r = −  Q r r   −1 = −