# Homework Help: Electric field in insulating slab

1. Feb 5, 2014

### temaire

A large, thin, insulating slab $2$ m x $2$ m x $5$ mm has a charge of $2$ x $10^{-10}$ C distributed uniformly throughout its volume. determine the electric field at observation point P, which is located within the slab, beneath its centre, $0.5$ mm from the top face.

I started off by making two Gaussian surfaces in the shape of cylinders. The first ranges from the bottom of the slab to point P. The second ranges from point P to the top of the slab. I then calculated the net electric field on point P due to the electric fields traveling out of the ends of the two Gaussian surfaces, using Gauss' Law.

$E = E_{1} - E_{2} = \frac{Q_{1}}{2A\epsilon_{o}} - \frac{Q_{2}}{2A\epsilon_{o}}$

where $E$ is the net electric field acting on point P, $E_{1}$ is the electric field from the first Gaussian surface, and $E_{2}$ is the electric field from the second Gaussian surface.

The charge density is as follows:

$\rho = Q/V = 2\cdot 10^{-10}/(2\cdot2\cdot0.005) = 1\cdot10^{-8}$

$Q_{1}=\rho V = \rho A \cdot 0.0045 = 4.5\cdot 10^{-11}A$

$Q_{2}=\rho V = \rho A \cdot 0.0005 = 5\cdot 10^{-12}A$

$E = \frac{4.5\cdot 10^{-11}A}{2A\epsilon_{o}} - \frac{5\cdot 10^{-12}A}{2A\epsilon_{o}}$

$E = 2.26$ N/C

My final answer is 2.26 N/C. Is my process correct?

2. Feb 5, 2014

### collinsmark

That looks correct to me.

Oh, except for one thing. You said, "Gaussian surfaces in the shape of cylinders." I think you meant pillboxes.