# Calculating the potential of a uniformily charged spherical solid

1. Sep 18, 2009

### noblegas

1. The problem statement, all variables and given/known data

Find the potential inside and outside uniformily charged spherical solid whose radius R and whose total charge is q.use infinity as your reference point

2. Relevant equations
$$V=-\int E* dl$$

gauss law = $$\int E *da=q/\epsilon_ 0$$
3. The attempt at a solution

This should be easy. Inside a solid sphere, E=0 so the potential inside sphere is zero. The electric field of a sphere is : $$E_sphere=(1/(4*\pi*\epsilon_ 0))*q/R^2$$ => $$V=-(1/(4*\pi*\epsilon_ 0))*q/R$$. Hmm... my solution is too easy; I know this solution was worked out in one of the examples found in my textbooks. Should I apply gauss law I take into account that $$dq=\rho*d\tau=\sigma*da$$ where $$d\tau=(4/3)*\pi R^3$$and$$da=4*\pi*r^2$$ ?

2. Sep 19, 2009

### Kalvarin

The electric field is only zero inside a conducting sphere. This is because all the charge migrates to the surface and it acts likes a hollow charged shell. In the question it does not say the sphere is conducting, it says it is uniformly charged. Therefore it must be acting as an insulating sphere, because a conducting sphere is not uniformly charged. Work it out assuming there is a field inside and you will get the right answer.

Also the potential inside a conducting sphere won't be zero, it will assume the value of the potential at the surface of the sphere.

3. Sep 19, 2009

### noblegas

I think I got it; for r>R, E(4*pi*r^2)=rho*(4/3*pi*R^3)/epilison_0 => and for R<r E(4*pi*r^2)==rho*(4/3*pi*R^3)/epilison_0

Last edited: Sep 19, 2009
4. Sep 20, 2009

### Kalvarin

Yep that looks right.