Question 3.8: An uncharged metal sphere of radius R is placed in an otherw...

The sphere is an equipotential—we may as well set it to zero. Then by symmetry the entire xy plane is at potential zero. This time, however, V does not go to zero at large z. In fact, far from the sphere the field is E_{0}\acute{z}, and hence

Since V = 0 in the equatorial plane, the constant C must be zero. Accordingly,the boundary conditions for this problem are

\begin{cases}(i) V = 0     when      r = R, \\ (ii) V →−E_{0}r  \cos \theta    for    r     \gg      R.\end{cases}                 (3.74)

We must fit these boundary conditions with a function of the form 3.65. The first condition yields

V(r, θ) =\sum\limits_{l=0}^{\infty }{\left(A_{l}r^{l}+\frac{B^{2l+1}}{r^{l+1}} \right)P_{l}(\cos\theta) }.                  (3.65)

or

B_{l}=-A_{l} R^{2l+1},          (3.75)

so

For r\gg R, the second term in parentheses is negligible, and therefore condition (ii) requires that

Evidently only one term is present: l = 1. In fact, since P_{1}(\cos \theta) = \cos \theta, we canread off immediately

A_{1} = −E_{0}, all other A_{l} ’s zero.

Conclusion:

V(r, θ) = −E_{0}\left(r-\frac{R^{3}}{r^{2}} \right)\cos \theta.       (3.76)

The first term (−E_{0}r \cos \theta) is due to the external field; the contribution attributable to the induced charge is

If you want to know the induced charge density, it can be calculated in the usual way:

\sigma (\theta )=-\epsilon _{0}\frac{\delta V}{\delta r}\mid _{r=R}= \epsilon _{0}E_{0}\left(1+2\frac{R^{3}}{r^{3}}\right)\cos\theta\mid _{r=R}=3\epsilon _{0}E_{0} \cos\theta.                           (3.77)

As expected, it is positive in the “northern” hemisphere (0 ≤ \theta ≤ \pi/2) and negative in the “southern” (\pi/2 ≤ \theta ≤ \pi).