# Electric potential on a non-uniform distribution - hollow sphere

I've been trying to solve this problem:

The electric potential on the surface of a hollow spherical shell of radius $$R$$ is $$V_0 cos\theta$$, where $$V_0$$ is a constant. In this problem we use spherical coordinates with origin at the center of the shell. What is the potential inside the shell?

Answer: $$V(r,\theta)=\frac{r}{R}V_0 cos\theta$$

I tried to find the charge distribution using the given potential but couldn't produce the correct result. Also, Gauss's Law doesn't help, as the electric flux is $$0$$ but we don't have any symmetry. Can someone please shine a light on this? Thanks in advance.

## 2 Answers

Since there is no charge inside the sphere, the potential satisfys the Laplace's Equation $$\nabla^2 V(r,\theta) = 0.$$

Due to the symmetry in the angle $$\phi$$, we can expand the potential in $$r$$ and Legendre function $$p_\ell(\cos\theta)$$:

$$V(r, \theta) = \sum_{n=0} a_n \frac{r^{n}}{R^{n+1}} P_n(\cos\theta).$$

Then match the boundary condition at $$r=R$$ to find the expansion coefficient $$a_n$$.

• Thanks! Watching some videos on YouTube to remember how to solve the Laplace Equation in polar coordinates. Apr 16, 2021 at 20:36
• Lapace Equation is solved by separation of variables, a very standard procedure.
– ytlu
Apr 17, 2021 at 4:09

This could either be a sphere in a uniform electric field or an electric dipole. If it is an electric dipole, the exterior voltage is $$V_e=V_0\frac{R^2}{r^2}\cos\left(\theta\right)$$ The charge density is given by $$\nabla\cdot\vec{D}=\rho$$ \begin{align} \nabla\cdot\vec{D}=& \,\varepsilon_0\left(\left(\frac{\partial V}{\partial r}\right)_{r=R}-\left(\frac{\partial V_e}{\partial r}\right)_{r=R}\right)\\ =& \, V_0\varepsilon_0\cos{\left(\theta\right)}\left(\frac{1}{R}\left(\frac{\partial r}{\partial r}\right)_{r=R}-R^2\left(\frac{\partial r^{-2}}{\partial r}\right)_{r=R}\right)\\ =& \,V_0\varepsilon_0\cos{\left(\theta\right)}\left(\frac{1}{R}\left(1\right)_{r=R}-R^2\left(-2r^{-3}\right)_{r=R}\right)\\ =& \,V_0\varepsilon_0\cos{\left(\theta\right)}\left(\frac{1}{R}+2\frac{1}{R}\right)\\ =&\,\frac{3V_0\varepsilon_0}{R}\cos{\left(\theta\right)} \end{align} So $$\rho=\frac{3V_0\varepsilon_0}{R}\cos{\left(\theta\right)}$$

• Thank you! It can't be an electric dipole, because there is nothing inside the sphere (I had tried the dipole and it led me to the wrong alternative). But considering a spherical shell inside an uniform field it worked! My question is, how did you see it had to be this exactly format? Apr 16, 2021 at 20:35
• @RodolfoM $z=r\cos(θ)$ As such, the voltage depends only on the z value and the dependence is linear. In other words, the internal field is uniform. I must say something though. Just because there's nothing in the sphere doesn't mean it isn't a dipole field. If you had a sphere whose surface charge density matched the one I calculated, it's internal field would be uniform but its external field would be that of a dipole. Apr 16, 2021 at 21:40