Question

What's wrong with my Gaussing?

Answer 1

Infinite uniformly negatively charged plate ( \[density= \sigma \] ) placed on top of a uniformly charged solid slab going infinitely sideways and with depth D.

Work out E field IN the slab.

1st Gauss surface: cylinder, cross-sectional area A, with both ends outside the system. Flux through the cylinder is \[ \frac{Q}{\epsilon_0} =\frac{\rho DA- \sigma A}{\epsilon_0} \], so the flux through the top end is\[ \frac{\rho DA- \sigma A}{2\epsilon_0} \]

2nd Gauss surface: shift the bottom end of the cylinder to be h above the bottom of the slab (i.e. still below much of the slab and the surface).
Flux through the surface is now \[ \frac{\rho (D-h)A- \sigma A}{\epsilon_0} \]
So flux through the bottom surface is [ \frac{\rho (D-h)A- \sigma A}{\epsilon_0}- \frac{\rho DA- \sigma A}{2\epsilon_0}\]
From this work out E field
\[ Flux=E \cdot A \] \[\frac{\rho (D-h)A- \sigma A}{\epsilon_0}- \frac{\rho DA- \sigma A}{2\epsilon_0}=EA \]
\[\frac{\rho (D-h)- \sigma }{\epsilon_0}- \frac{\rho D- \sigma }{2\epsilon_0}=E \]

Answer 2

\[ E= \frac{\rho D- 2\rho h-\sigma}{2\epsilon_0} outwards\]

Answer 3

I'm not so much interested with the answer or real method as much as where I went wrong.

Answer 4

you picked a surface that has different values of E on it.

Answer 5

Is this true?