# Deriving E of a charged spherical shell from V

1. Apr 5, 2014

### FallenLeibniz

Wrote my question up in Latex

Update: I have corrected the mistake when I stated the "textbook" version of the equation that the problem requires to be used. I have reposted the pdf with the same name.

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• ###### VvsE.pdf
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Last edited: Apr 5, 2014
2. Apr 5, 2014

### TSny

You need to do two integrations. One for finding the field at points outside the shell and the other for finding the field at points inside the shell. It looks like you have correctly done the one for points outside the shell. You should get a different result for points inside the shell.

3. Apr 5, 2014

### FallenLeibniz

I've taken another look at my integral I have set up. Now it seems as if it would work from a mathematical standpoint that for r<R, the seperation distance (i.e. little "r" in my integral) becomes equal to R which would give me a constant for the potential and therefore an E value of 0, however I don't see a justification for setting r=R within the integral on physical grounds.

Note: I have realized that I have miswritten the "overall" equation in my second section. I apologize and will correct it now.

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