# Potential of a dielectric sphere with radially dependent variable permittivity [closed]

Consider a dielectric sphere of radius $$R$$ in a constant external $$E$$-field, $$\mathbf{E}=E_0\mathbf{\hat{z}}$$, with a radially dependent variable permittivity $$\epsilon=\epsilon_0 (R/r)^2$$ for $$r.

I need to show that the potential satisfies the differential equation: $$\nabla^2V+\frac{d\ln{\epsilon}}{dr}\frac{\partial V}{\partial r}=0$$

I know how to begin (by substituting $$\mathbf{E}=-\nabla V$$ into $$\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$ to get Laplace's equation) and then writing $$\epsilon_0 =\epsilon(r/R)^2$$ to give: $$\nabla^2 V+\frac{d\ln{\epsilon}}{dr}\left(-\frac{\rho r}{2\epsilon_0}\right)=0$$

But I was wondering how I know that $$\frac{\partial V}{\partial r}=-\frac{\rho r}{2\epsilon_0}$$ considering that I haven't yet worked out what $$V$$ is?

The divergence of the electric flux density is given by $$\nabla \cdot \mathbf{D}= \nabla \cdot \varepsilon \mathbf{E} = \rho.$$ If $$\varepsilon$$ is not spatially constant, the divergence of the electric field is then given by: $$\epsilon \nabla\cdot \mathbf{E} + \mathbf{E}\cdot\nabla\varepsilon = \rho$$ which gives $$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon } - \mathbf{E} \cdot \nabla \ln\varepsilon$$
Plugging in the definition $$\mathbf{E} = -\nabla V$$ and since the volume contains no charges:
$$\nabla^2 V + \nabla V\cdot\nabla\ln\varepsilon = 0$$ Which in spherical coordinates is the differential equation in the question, assuming spherical symmetry.