# Uniformly polarized cylinder

1. Sep 25, 2006

### jdstokes

Hi all,

I'd like to get your opinion on the validity of my working for the following problem. According to my lecturer the answer is incorrect, but I can't figure out where I'm going wrong.

A cylinder of radius a is uniformly polarized in a direction perpendicular to the axis. The magnitude of the dipole moment per unit volume is P. Determine the electric field on the axis of the cylinder.

To me this seems like a straightforward application of Gauss' law, namely

$\int_{\mathrm{rectangle}}\vec{E} {\cdot} d \vec{A} + \int_{\textrm{half-cylinder}}\vec{E} {\cdot} d \vec{A} = \frac{2aP\ell}{\varepsilon_0}$

$E (2a \ell) + \int_{-\pi/2}^{+\pi/2}E\cos\theta a \ell d\theta = \frac{2aP\ell}{\varepsilon_0} \Rightarrow$

$E = \frac{P}{2\varepsilon_0}$

answer should by P / (4 epsilon).

Thanks

James

2. Sep 25, 2006

### Meir Achuz

Your use of Gauss is wrong.
Dimensionally, the answer must be proportional to P/epsilonnaught,
so any mistake will look close to right.
You have to integrate Coulomb's law for the bound surface charge.