解:(1)r<R,E=0,(2)r<Rpr 38 (3)中=5E6=E42=∑9 E0内 dv=4mr'dr d q= pdv= Ar'4rr'dr ∑q=4=4xzb ZAr Ear 1 元Ar R E 48 (r<R) 0解:(1) r R , E = 0 ,(2) r R , E r 0 3 = 2 E dS E4 r S = = (3) dV = r dr 2 4 dq = dV = Ar r dr 2 4 q dq Ar dr r = = 0 3 4 4 = Ar 4 0 2 1 E4 r Ar = 2 0 4 r A E = ( r R) r dr S r O R = 内 i q 0 1