# Potential inside of chrged non-conductive sphere

1. Oct 3, 2008

### anizet

1. The problem statement, all variables and given/known data
A charge is distributed uniformly throughout a non-conducting spherical volume of radius R. Show, that the potential at distance r from center (r<R) is given by:
V=q(3*R^2-r^2)/(8*PI*e0*R^3)

2. Relevant equations
From Gauss low:
E(r)=q*r/(4*PI*e0*R^3)

3. The attempt at a solution
After integrating I got:
V(r)=-q*R^2/(8*PI*e0*R^3)
which is in fact stupid, as it yields V=0 at r=0. So probably I integrated with wrong limits.
How to obtain correct formula?

2. Oct 3, 2008

### anizet

I made a small mistake in typing the formula on V that I obtained. It should be:
V(r)=-q*r^2/(8*PI*e0*R^3)

3. Oct 3, 2008

### Rake-MC

And R = radius of gaussian surface

4. Oct 3, 2008

### Rake-MC

Never mind, I've solved it.

You have two different functions for the electric field.
If r > R, $$E = \frac{kQ}{r^2}$$ where $$k = \frac{1}{4\pi\epsilon_0}$$
if r < R, $$E = \frac{kQr}{R^3}$$

so V of r < R will be the negative integral of r > R from R to infinity plus the negative integral of r < R from r to R.

$$V = - \int^{R}_{\infty} \frac{kQ}{r^2}dr - \int^{r}_{R} \frac{kQr}{R^3}dr$$

$$V = \frac{kR}{R} - \frac{kR}{2R^3}(r^2-R^2) = \frac{2kQR^2}{2R^3} + \frac{kQR^2}{2R^3} - \frac{kQr^2}{2R^3} = \frac{kQ(3R^2 - r^2)}{2R^3}$$

subbing k back in = $$\frac{Q(3R^2 - r^2)}{8 \pi \epsilon_0 R^3}$$

phew.. Hope there are no errors in that haha. I have been up for far too many consecutive hours..

Last edited: Oct 3, 2008
5. Oct 3, 2008

### anizet

Thank you very much for the reply!
I have just one more question: why should I integrate from infinity to zero, not from zero to infinity?

6. Oct 3, 2008

### anizet

Sorry again: from inf to r not from r to inf, I meant.

7. Oct 3, 2008

### anizet

One more correction:
Why should I integrate from infinity to r not from zero to r?
I hope, now it's ok...

8. Oct 3, 2008

### Rake-MC

because if we integrated from zero to r, that would give us V for inside the gaussian surface aswell as inside the sphere.

Sorry for the late reply I was away for sometime.