Dipole moment of non-uniform surface charged sphere

[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.

 

[dextercioby]

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

Daniel.

 

[meteorologist1]

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

Daniel.

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

 

[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

 

[meteorologist1]

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

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

 

[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)

 

[dextercioby]

Good answer...

Daniel.

 

[ehild]

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

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

 

[Majid]

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

 

[ehild]

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

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

 

[Majid]

ok, thank you.