# Finding the electric field in all regions given uniform charge occupying a<r<b

1. Feb 21, 2009

### Joan

Hi, I hope this is the appropriate place and format etc. One of the questions I have to answer is this:

Use Gauss's law to obtain the field, everywhere, of charge of uniform density roe, occupying the region a<r<b, where r is the distance from the origin.

I am completely flummoxed as to what assumptions I can make..

I am assuming that since we are asked to use Gauss that I should use a spherical symmetry..
I also think that where r<a the field should be zero, but can I assume that the charge lies on the inner surface only for this region?
And where r>b should I treat the whole dstribution as a point charge?
Also, I have started trying to find E inside the a<r<b, using Q= integral(roe dV),and I end up with a term involving b^3-a^3.

If anybody could tell me if i'm on the right track of give me any pointers it would be very much appreciated.

Many thanks.

Last edited: Feb 21, 2009
2. Feb 21, 2009

### clem

For r<a, Gauss shows that E=0.
For r>b, you aare right about the point charge.
Just use Gauss carefully for a<r<b.
Avoid integral rho dV.

3. Feb 21, 2009

### Joan

Hey thanks. In that case should I just leave integral (E.da) = rho/epsilon, ie leaving rho as is?

4. Feb 22, 2009

### clem

No. Read up on using Gauss's law in your textbook. You are not using it correctly.

5. Feb 22, 2009

### Joan

Hello, thankyou for your comments, I'm afraid i'm still stuck though...
I'm sure i'm being fairly thick here.
Gauss's law tells us that the flux through the chosen Gaussian surface (from the net electric field) is equal to the charge enclosed by the Gaussian surface divided by the permittivity of the material.
I suppose my question is where do I put my surfaces? I don't think it is OK either to assume that the charge resides on the two 'edges' (ie where r=a and r=b), as I am not told that the region is conducting. I think this would make the electric field zero in the region, due to the symmetry of the shells, is this right? If i integrate from a to b its no longer a surface but a volume....Can I do that? I really don't know where to go. (should I be working out the charge enclosed some other way to the one i suggested above? Would it be acceptable to find the fields for two solid spheres and then subtract the small from the large?) Sorry again for complete ineptitude and any help really is appreciated!

ps, apologies for continual edits...

Last edited: Feb 22, 2009
6. Mar 30, 2009

### CFDFEAGURU

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