# Spherically symmetric charge density given electric potential

1. Oct 9, 2012

### stauber28

1. A spherically symmetric charge distribution results in an electric potential of the form

What is the charge distribution?

2.
Hint: consider the difference in electric field between two values of r

Show that the answer is of the form

3. I have attempted several solutions but haven't gotten anywhere.

2. Oct 10, 2012

### Spinnor

Did you at least calculate the electric field?

E = - ∇ V where the gradient is in spherical coordinates.

3. Oct 10, 2012

### stauber28

I calculated E using the first equation. E=dV/dr

4. Oct 13, 2012

### stauber28

Correction E=-dv/dr <-- I am unsure to just have dr=dr or dr = (the derivative of everything in the 1st equations brackets[] ) - I decided to just have it equal dr

I then plugged the E value into the difference equation with r = r, and r' = (r+dr). I took the difference and set it equal to the right side of the same equation, and then solved for p(r). I hoped this would give me something in the form of charge density noted. However, I ended up with a large number of variables which would not simplify to this form.

Any Insights?

Last edited: Oct 13, 2012
5. Oct 13, 2012

### vela

Staff Emeritus
You have the right approach. You can neglect the higher-order terms. Keep only the terms proportional to dr.

6. Oct 13, 2012

### frogjg2003

You can have r'=0 and just integrate the right side of the equation from 0 to r. You now have Gauss' Law, and you can just solve algebraically for $\rho$.