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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.

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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 $a_n$.

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    $\begingroup$ Thanks! Watching some videos on YouTube to remember how to solve the Laplace Equation in polar coordinates. $\endgroup$ – RodolfoM Apr 16 at 20:36
  • $\begingroup$ Lapace Equation is solved by separation of variables, a very standard procedure. $\endgroup$ – ytlu Apr 17 at 4:09
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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 $$\rho=\frac{3V_0\varepsilon_0}{R}\cos{\left(\theta\right)}$$

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  • $\begingroup$ 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? $\endgroup$ – RodolfoM Apr 16 at 20:35
  • $\begingroup$ @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. $\endgroup$ – Laff70 Apr 16 at 21:40

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