# Dipole in a sphere

1. Jun 12, 2004

### Jonas Rist

Hi,

I have problems solving this:
Given: A sphere(Radius R) with a mathematical dipole in its center. I have to find the charge distribution on the sphere´s surface,

$$\sigma(r,\phi,\theta)$$

so that the resulting potential is zero for r>R. I think that
$$\sigma(r,\phi,\theta)=\sigma(\theta)$$
,but I have no idea how to find this special charge distribution(I suppose it is unique). I know the electric field of a dipole, but that doesn´t help much.
If anybody has an idea, please tell me!
Greets
Jonas

2. Jun 14, 2004

### turin

Superposition. There is a relationship between surface charge density and the normal component of the electric field that it generates. The dipole will also generate an electric field with a component normal to the sphere. You want the field of your surface charge to cancel the field for the dipole.

It is as you say: the surface charge density depends only on the polar angle.

To find the field due to the dipole, you can either look it up, or do a limit as a -> 0 and qa = constant, where a is the separation between two equal and opposite charges, q and -q.

3. Jun 14, 2004

### Jonas Rist

Hi,

thanks for the answer.
The normal component of the dipole field is
[TEX] E_r=\frac{qd 2 cos(\theta)}{4\pi\epsilon_0 r^3} [/TEX]
So I thought of choosing

[TEX] \sigma(\theta)=-\frac{qd 2 cos(\theta)}{4\pi R^3} [/TEX]

then the normal component should be cancelled out according to the following law:

[TEX] n(E_o-E_i)=\frac{\sigma}{\epsilon_0} [/TEX]

where E_o is the normal component outside(=0) and E_i the normal component inside. n is the unit normal vector.

But one professor of our university solved this problem too and he has the following result:

[TEX] \sigma(\theta)=-\frac{qd 3 cos(\theta)}{4\pi R^3} [/TEX]

He has a 3 instead of a 2, so my solution seems to be wrong.
So I´m still confused:(
Jonas

4. Jun 16, 2004

### reilly

The dipole moment of a sphere with radius R and a dipole charge distribution of C0=D0 * cos(theta) is (4pi/3)(R**3)D0. (this follows from the standard integral for a dipole moment) If this is to be equal and opposite to the original dipole moment, p0, then

D0 = p0 (3/4pi)/R**3

So the 3 is correct.
Regards,
Reilly Atkinson

5. Jun 16, 2004

### Jonas Rist

Ok, thanks!

I see now that my first way to solve the problem was wrong.
Ciao
Jonas

6. Jun 16, 2004

### reilly

Jonas -- Sorry , I fogot the minus sign in the sphere's dipole moment. RA

7. Jun 16, 2004

### Jonas Rist

No prob,
I got the idea of your solution, so I noticed that on my own.
Jonas