If now a conducting sphere of radius a is placed at the origin, the potenti will be that due to the charges +o at FR and their images +Qa/R at z +a/R Q4丌∈o Q4m∈o p=(2+R2+2rR COs 0)2(2+R2-2rR cos 8)/2 aO4丌∈ aQ4T∈ R R R Cos 8 where p has been expressed in terms of the spherical coordinates of the obser- vation point. In the first two terms R is much larger than r by assumption. Hence we can expand the radicals after factoring out R. Similarly, in the third and fourth terms, we can factor out rand then expand. The result is 20 a3 r cos 0 COS 4m∈oLR2 (213) where the omitted terms vanish in the limit r→∞. In that limit2g;me∈ok becomes the applied uniform field, so that the potential is cos e The first term(-Eoz) is, of course, just the potential of a uniform field Eo rcos 8+ 2Q 4TEo R RZr coS/+ The second is the potential due to the induced surface-charge density or, equiv alently, the image charges. Note that the image charges form a dipole of strength D=Qa/R X 2a/R=4Eo Eo a. The induced surface-charge density is =3∈ Eo cos0 We note that the surface integral of this charge density vanishes, so that there is no difference between a grounded and an insulated sphere 求半径为R0的导体球,置于均匀静电场E中,则求两半各收到电场给它的作用力,求此张 力大小。 tRO ER cos 8 p外=- EoR CoS6+q+求半径为 R0 的导体球，置于均匀静电场 E0 中，则求两半各收到电场给它的作用力，求此张 力大小。 0 0 0 0 4 Q R     = + 3 0 0 0 0 2 E cos -E R R Cos R     外 = + + 源场 偶极子项