#### The charged sphere

A spherical shell has spherical symmetry and because of this we write its potential as $V(R,\theta)=V_0(\theta)$ ($\theta$ is the “polar” angle of the spherical coordinate system).

a) When $V_0(\theta)=V_0(1-\theta^2)$ find the equation of definition for $V(R,\theta)$ if $r>R$.

b) When $V_0(\theta)=V_0(\cos^2 \theta +2)$ please evaluate the potential of this spherical shell at $r=0$.

#### The Spherical Shell Potential

The general expression for the potential of a charged sphere is found in Griffiths Electrodynamics, equation 3.65 from page 143. You can continue reading about this here…

$V(r,\theta)=\sum_{l=0}^{\infty} [A_lr^l+(B_l/r^{l+1})]*P_l(\cos \theta)$

There are two distinct regions, outside the shell and inside the shell.The first term in the equation refers to the region situated inside the sphere and the second term describes the region situated outside the sphere. Exactly on the shell when $r=R$ we have the limit (boundary) condition that gives the continuity of the solution.

a) As a result outside the sphere we write:

$V(r,\theta)=\sum_{l=0}^{\infty} (B_l/r^{l+1})*P_l(\cos \theta)$

where $B_l=[(2l+1)/2] * R^l\int_0^{\pi} V_0(\theta)P_l(\cos \theta)\sin \theta d\theta$

Because the limit condition is $V_0(\theta)=V_0(1-\theta^2)$ we have on the outside the potential:

$V(r,\theta)=\sum_l{[(2l+1)/2]R^l V_0*\int_0^{\pi}(1-\theta)^2 P_l(\cos\theta)\sin\theta*d\theta}*(1/r^{l+1})P_l(\cos\theta)$

One can not further simplify this equation because $V_0(\theta)$ can not be written as a sum of individual Legendre polynomials $P_l(\cos\theta)$. (An expression containg individual polynomials it would have been simpler).

b) Inside the sphere the first term that appears in the equation of potential is important.

$V(r,\theta)=\sum_{l=0}^{\infty} [A_lr^l]*P_l(\cos \theta)$

The potential has the coefficients equal to

$A_l=[(2 l+1)/(2R^l)]*\int_0^\pi V_0(\theta) P_l(\cos \theta)$

For the potential we write the boundary condition as a sum of Legendre polynomials $P_0$ and $P_2$.

$V_0(\theta)=V_0(\cos^2 \theta +2)=(2/3)P_2(\cos\theta)+(6/7)P_0(\cos \theta)$

We have to find the coefficients $A_2$ and $A_0$ from this equations. These coefficients are written as:

$A_2=V_0(5/2R^2)(2/3)*\int_0^\pi P_2^2(\cos\theta)\sin\theta *d\theta=(5/3R^2)[2/(2*2+1)]V_0=(2/3R^3)V_0$

$A_0=V_0(1/2)*\int_0^\pi P_0^2(\cos\theta)\sin\theta *d\theta=(1/2)[2/(2*0+1)]V_0=1$

As a result, inside the spherical shell we have the potential:

$V(r,\theta)=V_0[1+(2/3R^2)r^2P_2(\cos\theta)]$ and $V(0)=V_0$