Question 3.9: A specified charge density σ0(θ) is glued over the surface o...

You could, of course, do this by direct integration:

but separation of variables is often easier. For the interior region, we have

V(r,\theta)=\sum\limits_{l=0}^{\infty }{A_{l}r^{l}P_{l}(\cos\theta)}                (r\leq R)                       (3.78)

(no B_{l} terms—they blow up at the origin); in the exterior region

V(r,\theta)=\sum\limits_{l=0}^{\infty }{\frac{B_{l}}{r^{l+1} }P_{l}(\cos\theta)}            (r\leq R)                 (3.79)

(no A_{l} terms—they don’t go to zero at infinity). These two functions must be joined together by the appropriate boundary conditions at the surface itself. First, the potential is continuous at r = R (Eq. 2.34):

\begin{cases} (i) \quad\quad V_{in}=V_{out}, \quad at\quad r=R \\(ii) \quad\quad \epsilon \frac{\partial V_{in}}{\partial r}=\epsilon _0\frac{\partial V_{out}}{\partial r}, \quad at\quad r=R \\(iii) V_{out}\rightarrow -E_0rcos\theta \quad for\quad r\gg R \end{cases}              (2.34)

Potentials

\sum\limits_{l=0}^{\infty }{A_{l}R^{l}P_{l} (\cos\theta)}=\sum\limits_{l=0}^{\infty }{\frac{B_{l}}{R^{l+1}} P_{l}(\cos\theta)}                             (3.80)

It follows that the coefficients of like Legendre polynomials are equal:

B_{l}=A_{l}R^{2l+1}.                        (3.81)

(To prove that formally, multiply both sides of Eq. 3.80 by P_{\acute{l}} (\cos \theta) \sin \theta and integrate from 0 to π, using the orthogonality relation 3.68.) Second, the radial derivative of V suffers a discontinuity at the surface (Eq. 2.36):

\frac{\delta V_{above}}{\delta n}-\frac{\delta V_{below}}{\delta n} =\frac{1}{\epsilon_{0}}\sigma         (2.36)

\left(\frac{\delta V_{out}}{\delta r}-\frac{\delta V_{in}}{\delta r}\right)\mid _{r=R} =-\frac{1}{\epsilon_{0}}\sigma_{0}(\theta)        (3.82)

Thus

or, using Eq. 3.81,

\sum\limits_{l=0}^{\infty }(2l+1)A_{l}R^{l-1}P_{l}(\cos\theta)=\frac{1}{\epsilon _{0}}\sigma _{0}(\theta)               (3.83)

From here, the coefficients can be determined using Fourier’s trick:

A_l=\frac{1}{2\epsilon _0R^{l-1}}\int_{0}^{\pi}{\sigma _0(\theta )P_l(cos\theta )sin\theta d\theta .}              (3.84)

Equations 3.78 and 3.79 constitute the solution to our problem, with the coefficients given by Eqs. 3.81 and 3.84.
For instance, if

\sigma _0(\theta )=k\cos\theta =kP_1(cos\theta )               (3.85)

for some constant k, then all the A_l’s are zero except for l = 1, and

The potential inside the sphere is therefore

V(r,\theta )=\frac{k}{3\epsilon _0}rcos\theta \quad (r\leq R),                (3.86)

whereas outside the sphere

V(r,\theta )=\frac{kR^3}{3\epsilon _0}\frac{1}{r^2} cos\theta \quad (r\geq R),             (3.87)

In particular, if σ_0(θ) is the induced charge on a metal sphere in an external
field E_0\widehat{z} so that k = 3\epsilon _0E_0 (Eq. 3.77), then the potential inside is E_0rcosθ=E_0z, and the field is -E_0\widehat{z} —exactly right to cancel off the external field, as of course it should be. Outside the sphere the potential due to this surface charge is

\sigma (\theta )=-\epsilon _0\frac{\partial V}{\partial r} \mid_{r=R}=\epsilon _0E_0\left(1+2\frac{R^3}{r^3} \right) cos\theta \mid_{r=R}=3\epsilon _0E_0cos\theta          (3.77)

consistent with our conclusion in Ex. 3.8.