# Dipole moment of given charge distribution

1. May 30, 2014

### ks_wann

So, I've got a charge distribution given by:

\rho(r,\phi,z)=\frac{q}{2\pi R}\delta(r-R)\delta(z)\cos(2\phi)

This, if I'm not mistaken, translates into a circular charge distribution located in the z-plane, a distance R from origo.

Thus

\rho(R,\phi,0)=\frac{q}{2\pi R}\cos(2\phi)

I compute the total charge of the ring, which by my calculations is 0. Now I want to use

\mathbf{p}=\int\mathbf{r^{\prime}}\rho(\mathbf{r^{\prime}})dl^{\prime}

to find the dipole moment, where

dl^{\prime}=Rd\phi.

Inserting in the equation, I get:

\mathbf{p}=\int R\mathbf{\hat{r}}\frac{q}{2\pi R}\cos(2\phi)Rd\phi \\ = \frac{Rq}{2\pi}\mathbf{\hat{r}}\int_{0}^{2\pi} \cos(2\phi)d\phi = 0

I'm not sure if the last steps makes sense..

I'd be grateful for any help.

2. May 30, 2014

### rude man

... a circular charge distribution located in the z=0 plane, a distance R from the origin.

Right.

What you did was fine as far as it went. But I would now split up your p integral into x, y and z components: px, py and pz.

So we would then get
p = px i + py j + pz k and

r' = R cosø i + R sinø j + 0 k
most of which you already have.

So take your integral for p, separate into x, y & z components, and finally obtain p from px, py and pz.

I haven't finished solving the problem but will if you are interested.

Last edited: May 31, 2014
3. May 30, 2014

### dauto

The answer is indeed zero by symmetry but you made a mistake in your calculation. the unit vector "r-hat" cannot be removed from the integral because its direction is a function of ø.

4. May 31, 2014

### rude man

I did the integrations and confirm that px = py = pz = 0.

5. May 31, 2014

### ks_wann

I did the integration in x, y and z components, and I also get 0.

Is it much harder to solve the problem in cylindrical coordinates, given that "r-hat" is a function of ø? I would include the "r-hat" into the integral in my final step, find the expression for "r-hat" of ø (if one such exists in cylindrical coordinates?) and integrate.

Also, is the dipole moment 0 due to no bound charges in the distribution?

6. May 31, 2014

### rude man

I haven't tried to integrate in cylindrical coordinates so don't know. I might look into it.

All I can tell you regarding net charge is that if there is no net charge the coordinate system can be chosen arbitrarily. Naturally in this case the origin should be the center of the circular charge distribution.

7. May 31, 2014

### rude man

I guess I have a basic prblem with cylindrical coordinates. It's based on the fact that a vector gives displacement but not position. Example:
r = 2 i + 3 j
displacement = √(4+9) = √13.
position is (2,3).

But, in cylindrical coordinates using unit vectors, the same r is r = √13 r-hat + 0
The displacement is correct but the position can be anywhere on a circle of radius √13 .

I usually wind up with going x = r cosθ, y = r sinθ, z. In other words, back to cartesian.

Of course you could define this vector as √13 ∠ tan-1(3/2) but that is not using unit vectors.

I would welcome comments from anyone.