chapter6（精选）

1)[Q/(4pe0er(R+d)2]4p(R+d)2Р=(er-1)Q/erР2.球形电容器 C=4pe0RРQ1=C1V1= 4pe0RV1 Q2=C2V2= 4pe0RV2РW0=C1V12/2+C2V22/2=2pe0R (V12+V22)Р两导体相连后 C=C1+C2=8pe0RРQ=Q1+Q2= C1V1+C2V2=4pe0R(V1+V2)РW=Q2/(2C)= [4pe0R(V1+V2)]2/(16pe0R)Р=pe0R(V1+V2)2Р静电力作功Р A=W0-WР=2pe0R (V12+V22)-pe0R(V1+V2)2Р=pe0R(V1-V2)2Р=1.11×10-7JР练习8 静电场习题课Р一、选择题 D B A C AР二、填空题Р1. 9.42×103N/C, 5×10-9C.Р2. .Р3 R1/R2, 4pe0(R1+R2), R2/R1.Р三、计算题Р1. (1)拉开前 C0=e0S/dРW0=Q2/(2C0)= Q2d/(2e0S)Р拉开后 C=e0S/(2d)РW=Q2/(2C)=Q2d/(e0S)РDW=W-W0= Q2d/(2e0S)Р(2)外力所作功РA=-Ae=-(W0-W)= W-W0= Q2d/(2e0S)Р外力作功转换成电场的能量Р{用定义式解:A==Fd=QE¢dР=Q[(Q/S)/(2e0)]d= Q2d/(2e0S) }Р2. 洞很细,可认为电荷与电场仍为球对称,由高斯定理可得球体内的电场为РE=(r4pr3/3)/(4pe0r2)(r/r)Р=rr/(3e0)=Qr/(4pe0R3)РF=-qE=-qQr/(4pe0R3)РF为恢复力, 点电荷作谐振动Р-qQr/(4pe0R3)=md2r/dt2Рw=[ qQ/(4pe0mR3)]1/2Р因t=0时, r0=a, v0=0,得谐振动A=a,j0=0故点电荷的运动方程为