(2)由边界条件确定Q′和〃 设Oo=br=√R2+a2-2 Ra cos e r+b-2rb cos e 0 0 O R=Ro R=Ro R=Ro Q-(R662)-20-Rbcos8=2(ro+a)-20 cos e 因O任意的Q2b=Q2aQ(R2+b2)=Q2(R2+a2) R 解得①b RoO 2 RO ②b=aQ=±Q2 cos 2 2 r = R + a − Ra 2 cos 2 2 r  = R + b − Rb 0 0 0 0 2 2 2 2 0 0 R R R R R R R r Q r Q r Q r Q = = =   = =    = + （2）由边界条件确定 Q 和 r  设 OQ = b 2 2 2 2 0 0 Q R b Q R b ( ) 2 cos + −  2 2 2 2 0 0 = + − Q R a Q R a   ( ) 2 cos 因  任意的 Q b Q a 2 2 =  ( ) ( ) 2 2 0 2 2 2 0 2 Q R + b = Q R + a 解得 ① a R Q Q a R b 0 2 0 =  =  ② b = a Q = Q a R Q Q a R b 0 2 0 =  = − 机动 目录 上页 下页 返回 结束