# Dipole moment of non-uniform surface charged sphere

1. Jan 27, 2005

### meteorologist1

I'm trying to find the dipole moment of a non-uniform surface charge distribution on a sphere of radius a:

The surface charge distribution is:
$$\sigma = \sigma_{0} cos \theta$$
where theta is the polar angle.

Here is what I did:
$$\vec{p} = \int\vec{r}\sigma dA$$
$$= \int r \sigma_{0} cos \theta (2\pi r dr d\theta)$$

and I'm thinking r should be integrated from 0 to a and theta integrated from -pi/2 to pi/2, but I'm not sure. Please help; thanks.

2. Jan 27, 2005

### dextercioby

Hold on.
What is the surface element in spherical coordinates??

Daniel.

3. Jan 27, 2005

### meteorologist1

Ok I see. Should I use: $$r dr d\theta$$ ?

4. Jan 27, 2005

### vincentchan

dA is not $2 \pi r dr d \theta$, how could the surface element depend on r ???

try $$dA = 2 \pi R^2 sin \theta d \theta$$
and go do some reading on 3D integral

5. Jan 27, 2005

### meteorologist1

I get 0 for the answer if I do it like you said:

$$\int a \sigma_{0} cos\theta (2\pi a^2 sin\theta d\theta)$$
$$= 2\pi a^3 \sigma_{0} \int sin\theta cos\theta d\theta = 0$$ theta ranges from -pi/2 to pi/2

6. Jan 27, 2005

### vincentchan

people measure theta from 0 to pi

if your question define theta like you said, you will get zero anyway because the surface charge is symmetric....(save you some time to do the integral)

7. Jan 27, 2005

### dextercioby

Daniel.

8. Jan 29, 2005

### ehild

You miss the z component of the position vector from your integrand and you have to integral from 0 to pi. Theta is the angle of the position vector with respect to the z axis.

$$p_z=\int_{0}^{\pi}{\sigma* z* dA } = \int_{0}^{\pi}{(\sigma_0 \cos{\theta})*(acos{\theta})*(2\pi a^2 \sin{\theta}d\theta)}$$

ehild

9. Jun 20, 2005

### Majid

may you explain it. I didn't get why z component is here.

10. Jun 21, 2005

### ehild

Remember the definition of dipole moment.

$$\vec{p} = \int {\sigma \vec{r} dA }$$

Because of symmetry, the dipole moment has got only z component here. The z component of the eq. above is:

$$p_z = \int {\sigma z dA}$$

ehild

11. Jun 21, 2005

### Majid

ok, thank you.