A recap on Gauss' Law

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Gauss' Law is a topic that we covered in Physics-C that was pretty much completely new to us. We didn't talk about it about all in Physics-B, so it's one of the harder things we have to do in Physics-C this year.

Because of this, for myself and for others, I want to type up some of the derrivations and notes so that its fresh in my mind and can maybe help some other people who need it!

The equation to remember:

$png.latex? \phi =\oint \underset{E}{\rig$

When you have a conducting sphere with uniform charge Q and radius R:

1. r < R

$png.latex? \oint \underset{E}{\rightarro$

$png.latex? E\oint \underset{dA}{\rightar$

$png.latex? E(4\pi r^{2})=\frac{0}{\epsil$

$png.latex? E=0$

2. r > R

$png.latex? \oint \underset{E}{\rightarro$

$png.latex? E(4\pi r^{2})=\frac{Q}{\epsil$

$png.latex? E=\frac{Q}{\epsilon _{0}(4\pi$

$png.latex? E=\frac{kQ}{r^{2}}$

When you have an infinite plane with charge density of $png.latex?\sigma$

The First step is to assume that there is a finite cylinder that comes out of either side of the infinite plane. This will help with the derrivation and any dimmensions of the cylinder will later cancel.

$png.latex? \phi =\oint \underset{E}{\rig$

$png.latex? \sigma =\frac{Q}{A}$

$png.latex? \phi _{top}+\phi _{bottom}=\f$

$png.latex? EA+EA=\frac{\sigma A}{\epsilo$

$png.latex? 2E=\frac{\sigma }{\epsilon _{$

$png.latex? E=\frac{\sigma }{2\epsilon _{$

For 2 infinite planes that are uniformly charged, one is $png.latex? +\sigma$ and the other is $png.latex? -\sigma$ , the electric fields point in the same direction, so you have: