Gauss's Law on a Charged Sphere

[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.

Homework Statement

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.

Homework Equations

Gauss's Law.

E = -\nabla V

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.

 

[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

 

[LeePhilip01]

Got it all sorted now! Thanks!