# Gauss's Law on a Charged Sphere

1. Aug 7, 2011

### LeePhilip01

Hi I'm sorry if this is posted in the wrong section or it's laid out wrong but I have a question that I need a bit of help with.

1. The problem statement, all variables and given/known data
I'm given: A spherically symmetrical charge distribution is contained within a sphere of radius a with no charge outside. At a distance r (r $\leq$ a) from the centre of the sphere the potential may be written as:

V = -Ar$^{3}$ + B

Then I'm asked to calculate a bunch of things which hopefully I've got right.

(a) Calculate the electric field intensity for r < a.

(b) What is the electric field intensity for r = a.

(c) Using Gauss's law calculate the total charge in the sphere.

(d) What is the electric field for r > a?

(e) What is the potential for r > a if V$_{\infty}$ = 0?

(f) Show that the difference in potential between r = a/2 and r = 2a is given by (19/8)Aa$^{3}$.

It's the last one I'm stuck on.

2. Relevant equations
Gauss's Law.

E = -$\nabla$ V

3. The attempt at a solution

Ok. So for (a) I differentiated and got E = 3Ar$^{2}$

(b) I just subbed r=a and got E = 3Aa$^{2}$

(c) I got that E for r=a with Gauss's Law: E = $\frac{Q}{4\pi\epsilon_{0}a^{2}}$. Then equated that with answer from (b) to get Q = 12$\pi\epsilon_{0}$Aa$^{4}$

(d) I used Gauss's Law again to obtain: E = $\frac{Q}{4\pi\epsilon_{0}r^{2}}$.

Then I subbed in Q from part (c) to get: E = 3A$\frac{a^{4}}{r^{2}}$.

(e) I integrated -E with boundaries r and 0 and got: V = 3A$\frac{a^{4}}{r}$.

(f) Now I subbed r=a/2 into the equation for V given in the question. And r=2a into the equation I just got in part (e). But cannot seem to get (19/8)Aa$^{3}$ at all.

Thanks for any help.

2. Aug 7, 2011

### ehild

For e, you need to integrate between r and infinity as the potential is zero at infinity.
For f, you need to match B so as V=-Ar3+B be the same at r=a as the outer potential.

ehild

Last edited: Aug 7, 2011
3. Aug 7, 2011

### LeePhilip01

Got it all sorted now! Thanks!