Homework Help: 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.