# Potential difference of concentric conducting shells

1. Sep 27, 2011

### faint545

My professor said to in order to solve this, integrate the electric field to find the electric potential...

$\Delta V = -\int\stackrel{\rightarrow}{E}dl$

My question is, using Gauss's Law, ($\oint E_{n}dA = \frac{Q}{\epsilon}$), how do I go about finding Q?

Isn't Q just the charge of the shell?

2. Sep 27, 2011

### G01

In Gauss's law, Q is the charge enclosed by your Gaussian surface. So, first decide where you Gaussian surface will be, then add up all the charge inside of it.

HINT: You want your surface to be in the region where you want to find the electric field.

3. Sep 27, 2011

### faint545

This is what I have drawn (see attachment). Is the basic idea here to integrate the electric field of the outer Gaussian surface from b to a? If so, what about the inner Gaussian surface?

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4. Sep 28, 2011

### G01

You want to find the potential difference between the shells, so you don't need the Gaussian surface outside the larger shell.

Try this: Take your inner surface and place it at an arbitrary point r. Then, find E using the standard approach when using Gauss's law. You will then have E between the plates as a function of r. Can you use that to find the potential difference between the plates?

5. Sep 28, 2011

### faint545

Alright... see attached image.

Now, how do i determine if i should integrate from a to b or from b to a?

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• ###### 2011-09-28_12-25-07_614.jpg
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6. Sep 28, 2011

### G01

That all looks good.

As far as your integration bounds are concerned, it doesn't matter. The sign of your end result will change, but that's just like hooking up a voltmeter in reverse: You will still get the right pot. difference, just the sign will change.

7. Sep 29, 2011

### faint545

thanks for your guidance

8. Sep 29, 2011

No problem!