# Dielectric cylinder in uniform electric field: nonlong cylinder

Problem of dielectric cylinder in uniform electric field is well known. For example, Jackson textbook or Griffiths textbook or online solution here.

Solution always given for case of long cylinder. So that $z$ dependence is removed from solution of Laplace equation $\nabla^2 \phi = 0$ in cylindrical coordinates $(r, \theta, z)$. Solution in form $\phi = \phi (r, \theta)$ is given here.

Now consider the case of nonlong cylinder. What are the additional boundary conditions on potential $\phi = \phi (r, \theta, z)$, compared to those of long cylinder case? Or, how to treat end effects?

This problem can be solved only numerically for example with integral equation. van Bladel in his Electromagnetic Fields derives the so-called Phillips Integral Equation (Equation 3.152, in the 2nd edition) for points $\mathbf{r}\in S$ on the surface of the object$$\phi ^i (\mathbf{r})= \frac{\epsilon_r+1}{2} +\frac{\epsilon_r-1}{2} + \int_S \phi({\mathbf{r}}')\frac{\partial } {\partial {n}'} \left ( \frac{1}{\left | \mathbf{r} - \mathbf{r}' \right |} \right ) {dS}'$$ Here $\phi ^i (\mathbf{r})$ is the external potential in which the homogeneous object of surface $S$ is immersed. Later van Bladel shows how to use this for a cube. Once you know the potential on the surface more conventional numerical methods are applicable. One can of course discretize Poisson's equation directly for the whole space, too.