0
$\begingroup$

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 $r<R$.

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.