如图所示,接地的空心导体球壳内半径为R,在空腔内一直径上的P1和P2处...
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发布时间:2024-10-06 13:19
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时间:2024-10-26 16:06
(1)如图1所示,S为原空腔内表面所在位置,q1′的位置应位于.OP1的延长线上的某点B1处,q2′的位置应位于.OP2的延长线上的某点B2处.设A1为S面上的任意一点,根据题意有
kq1A1P1+kq1′A1B1=0…①kq2A1P2+kq2′A1B2=0…②
怎样才能使 (1)式成立呢?下面分析图1中△OP1A1与△0A1P1的关系.
若等效电荷q1′的位置B1使下式成立,即
.OP1?.OB1=R2…③
即.OP1.OA1=.OA1.OB1…④
则△OP1A1∽△OA1B1
有.A1P1.A1B1=.OP1.OA1=aR…⑤
由 ①式和 ②式便可求得等效电荷q1′,
q1′=?Raq1…⑥
由 (3)式知,等效电荷q1′的位置B1到原球壳中心位置O的距离
.OB1=R2a…⑦
同理,B2的位置应使△OP2A1∽△OA1B2,用类似的方法可求得等效电荷
q2′=?Raq2…⑧
等效电荷q2′的位置B2到原球壳中心O位置的距离
.OB2=R2a…⑨
(2)A点的位置如图2所示.A的电势由q1、q1′、q2、q2′共同产生,即
UA=kq(1.P1A?Ra1.B1A+1.P2A?Ra1.B2A)…⑩
因
.P1A=r2?2racosθ+a2
.B1A=r2?2r(R2a)cosθ+(R2a)2,
.P2A=r2+2racosθ+a2.
.B2A=r2+2r(R2a)cosθ+(R2a)2,
代入 ⑩式得:
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