Question 3.58: Find the charge density σ(θ) on the surface of a sphere (rad...

Potential of q:  V_{q}( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{q}{ᴫ}   , where  ᴫ^{2}=a^{2}+r^{2}-2 a r \cos \theta .

Equation 3.94,

\frac{1}{ᴫ}=\frac{1}{r} \sum_{n=0}^{\infty}\left(\frac{r^{\prime}}{r}\right)^{n} P_{n}(\cos \alpha)                      (3.94)

with r^{\prime} \rightarrow a \text { and } \alpha \rightarrow \theta: \quad \frac{1}{ᴫ}=\frac{1}{r} \sum_{n=0}^{\infty}\left(\frac{a}{r}\right)^{n} P_{n}(\cos \theta) . So 

V_{q}(r, \theta)=\frac{q}{4 \pi \epsilon_{0}} \frac{1}{r} \sum_{n=0}^{\infty}\left(\frac{a}{r}\right)^{n} P_{n}(\cos \theta) .

Meanwhile, the potential of σ is (Eq. 3.79)

V_{\sigma}(r, \theta)=\sum_{l=0}^{\infty} \frac{B_{l}}{r^{l+1}} P_{l}(\cos \theta) .

Comparing the two \left(V_{q}=V_{\sigma}\right) \text { we see that } B_{l}=\left(q / 4 \pi \epsilon_{0}\right) a^{l}, \text { and hence (Eq. 3.81) } A_{l}=\left(q / 4 \pi \epsilon_{0}\right) a^{l} / R^{2 l+1} . Then (Eq. 3.83)

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

\sum_{l=0}^{\infty}(2 l+1) A_{l} R^{l-1} P_{l}(\cos \theta)=\frac{1}{\epsilon_{0}} \sigma_{0}(\theta)                   (3.83)

\sigma(\theta)=\frac{q}{4 \pi R^{2}} \sum_{l=0}^{\infty}(2 l+1)\left(\frac{a}{R}\right)^{l} P_{l}(\cos \theta)=\frac{q}{4 \pi R^{2}}\left[2 \sum_{l=0}^{\infty} l\left(\frac{a}{R}\right)^{l} P_{l}(\cos \theta)+\sum_{l=0}^{\infty}\left(\frac{a}{R}\right)^{l} P_{l}(\cos \theta)\right].

Now (second line above, with r → R)

\frac{1}{\sqrt{a^{2}+R^{2}-2 a R \cos \theta}}=\frac{1}{R} \sum_{l=0}^{\infty}\left(\frac{a}{R}\right)^{l} P_{l}(\cos \theta) .

Di↵erentiating with respect to a:

\frac{d}{d a}\left(\frac{1}{\sqrt{a^{2}+R^{2}-2 a R \cos \theta}}\right)=-\frac{(a-R \cos \theta)}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}}=\frac{1}{a R} \sum_{l=0}^{\infty} l\left(\frac{a}{R}\right)^{l} P_{l}(\cos \theta) .

Thus

=\frac{q}{4 \pi R} \frac{\left[-2 a(a-R \cos \theta)+\left(a^{2}+R^{2}-2 a R \cos \theta\right)\right]}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}}=\frac{q}{4 \pi R} \frac{\left(R^{2}-a^{2}\right)}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} .