?2??2????0即2?2?0?x?y2边界条件为?x?0,0?y?b?0?x?a,0?y?b?U0?y?0,0?x?a?0?y?b,0?x?a?0分离变量,设??f(x)g(y)代入方程并且两边同时除以f(x)g(y)得f''(x)g''(y)??0f(x)g(y)f''(x)g''(y)设??则=??f(x)g(y)方程可写成以下形式f''(x)??f(x)?0(1)g''(y)??g(y)?0(2)
解方程(2)并要求满足边界条件
?y?0,0?x?a?0?y?b,0?x?a?0得只有??0时方程满足要求n2?2解得?n?2bgn(y)?sin(
n?y) b将代入方程(1),并满足边界条件?x?0,0?y?b?0n?x)bn?n?则?n?Ansinh(x)sin(y)
bb解得fn(x)?Ansinh(则电位的通解为???Ansinh(n?1?
n?n?x)sin(y) bb
代入边界条件?x?a,0?y?b?U得?x?a??Ansinh(n?n?a)sin(y)?Ubbn?1m?两边同时乘以sin(y)并对y从0到b积分,并由bbn?m?n?m时?sin(y)sin(y)dy?00bbbn?m?bn?m时?sin(y)sin(y)dy?得0bb2b?n?n?Asinh(a)sin(y)dy?n?0n?1bb?bn?n???An?sinh(a)sin(y)dy0bbn?1bm??Amsinh(a)2bbm? ??Usin(y)dy(3)0b?(1)U=U0时,由方程(3)得bbm? U0(1?cosm?)?Amsinh(a)m?2b1(m?1,3,5K)m?sinh(a)b 代入电位的通解求得电位为n?sinh(x)?4U0n?b???sin(y)n?bn?1,3,5Kn?sinh(a)b则Am?(2)U?U0sinb4U0m??yb时bm??ym?Usin(y)dy?Usinsin(y)dy ?0?00bbbb=U0(m=1)2
由方程(3)得bb?U0?A1sinh(a)22bU0则A1?
?sinh(a)b代入电位的通解求得电位为?xsinh()bsin(?y)??U0?bsinh(a)b3.9一个沿+y方向无限长的导体槽，其底面保持电位为U0，其余两面的电位为零，如图3.9所示。求槽内的电位函数。
电位分布满足拉普拉斯方程?2??2????0即2?2?0?x?y2
边界条件为?x?0?0?x?a?0?y?0?U0?y???C分离变量,设??f(x)g(y)代入方程并且两边同时除以f(x)g(y)得f''(x)g''(y)??0f(x)g(y)f''(x)g''(y)设???则=?f(x)g(y)方程可写成以下形式f''(x)??f(x)?0(1)g''(y)??g(y)?0(2)
解方程(1)并要求满足边界条件?x?0?0?x?a?0得
只有??0时方程满足要求n2?2n?解得?n?2fn(x)?sin(x)aa将代入方程(2),并满足边界条件?y???C解得gn(y)?Ane?n?yan?y
?n?则?n?Ansin(x)eaa
则电位的通解为?n????Ansin(x)eaan?1?n?y代入边界条件?y?0?U0得U0??Ansin(n?x)an?1m?两边同时乘以sin(x)并对x从0到a积分,并由aan?m?am?n时?sin(x)sin(x)dx?Am0aa2an?m?m?n时?sin(x)sin(x)dx?0得0aaaa?m?n?m?aUsin(x)=Asin(x)sin(x)dx?Amn?00?0?aaa2n?1aa既U0(1?cosm?)?Amm?24U0则Am?(m?1,3,5K)m?
代入电位的通解方程得???n?1,3,5K??n??y4U0n?sin(x)ean?a4.3若半径为a、电流为I的无线长圆柱导体置于空气中，已知导体的磁导率为?0，求导体内、外的磁场强度H和磁通密度B。
解：（1）导体内：0?? l???'I?2I?2II?2H2???.??H? I== 所以，， ，1.1a2a2?a22?a2'