# Electric field of dipole moment

1. Dec 17, 2015

### Incand

1. The problem statement, all variables and given/known data
A solid sphere, radius $R$,is centered at the origin. The "northern" hemisphere carries a uniform charge density $\rho_0$ and the "southern" hemisphere a uniform charge density $-\rho_0$. Find the approximate field $\mathbf E(r,\theta)$ for points far from the sphere ($r >> R$).

2. Relevant equations
For a dipole where $\mathbf p$ is located at the origen and points in $z$ direction the electric field is given by
$\mathbf E_{\text{dip}}(r,\theta) = \frac{p}{4\pi \epsilon r^3}(2\cos \theta \hat r + \sin \theta \hat \theta)$.

Dipole moment
$\mathbf p = \int \mathbf r' \rho(\mathbf r') d\tau'$.

3. The attempt at a solution
I think I solved this one but I'm not entirely sure so I'm hoping someone could take a look and see if I'm doing this roughly right.

The dipole has to point in the $\hat z$ direction since symmetry. Our $\mathbf r'$ vector should then be $\mathbf r' = R\cos \theta \hat z$. So the dipole moment is
$\mathbf p = \hat z \int_S R\cos \cdot R^2\sin \theta d\theta d\phi = \pi R^3\hat z\left[ \int_0^{\pi/2} \sin 2\theta d\theta - \int_{\pi/2}^\pi \sin 2\theta d\theta \right] = 2\pi R^3\hat z$
using that the area element of an $r=R$ surface is $R^2\sin \theta d\theta d\phi$.
Our electric field is then (only taking account of the dipole term in the multipole expansion)
$\mathbf E(r,\theta) =\frac{R^3}{2\epsilon_0 r^3}(2\cos \theta \hat r +\sin \theta \hat \theta)$.
Does this look correct? It's certainly the field of a dipole and it's stronger when $\theta=0$ than when $\theta=\pi/2$ which makes sense but I don't know how to check it really.

2. Dec 17, 2015

### BvU

It doesn't have the right dimension yet....
I also don't see an integration over the volume, only the shell ?

Last edited: Dec 17, 2015
3. Dec 17, 2015

### Incand

Right, I dropped the charge density $\rho$, I was supposed to move that outside the integral and it disappeared.
So the final answer should be $\mathbf E(r,\theta) =\frac{R^3\rho}{2\epsilon_pr^3}(2\cos \theta \hat r + \sin \theta \hat \theta)$.

This should have the right dimension. the $R^3/r^3$ units cancel out. The unit of $\rho$ is $C/m^2$ and the unit of $\epsilon_0$ is $C/(Vm)$. So combining those we have $C/m^2\cdot \frac{Vm}{C} = V/m$ volts per meter which is the unit for the electric field.

4. Dec 17, 2015

### BvU

Still not happy.
The dimension of $\rho$ is C/m3 (the sphere is solid -- it'not a shell !)

5. Dec 17, 2015

### Incand

I need to get better at reading questions, I thought I had a shell with surface charge. But perhaps I solved it for a shell with surface charge $\rho$!

Lets redo the calculations, now with volume integrals.
The dipole should still be along the $z$-axis but now with $\mathbf r' = r\cos \theta \hat z$. and the volume element is $d\tau' = r^2\sin \theta$.
$\mathbf p = \int_V \mathbf r' \rho(\mathbf r') d\tau' = \hat z \rho \int_0^{2\pi}d\phi \int_0^R r^3dr\left[\int_0^{\pi/2}\sin\theta \cos \theta d\theta - \int_{\pi/2}^\pi \sin \theta \cos \theta d\theta \right] = \frac{\pi \rho R^4\hat z}{2}\left[\frac{-cos 2\theta}{4}\bigg|_0^{\pi/2} + \frac{\cos2 \theta}{4}\bigg|_{\pi/2}^\pi \right] = \frac{\pi \rho R^4\hat z}{2}$
Then the field is
$\mathbf E = \frac{\pi \rho R^4 \hat z}{2\cdot 4\pi \epsilon_0 r^3} (2\cos \theta \hat r + \sin \theta \hat \theta) = \frac{\rho R^4 \hat z}{8 \epsilon_0 r^3}(2\cos \theta \hat r + \sin \theta \hat \theta).$
with units $\frac{C}{m^3}\frac{m^4}{m^3}\frac{1}{C/(Vm)} = \frac{V}{m}$.

6. Dec 17, 2015

### BvU

I think so, yes. Note that the upper half has charge $\ q = {2\over 3}\rho\pi R^3\$ so you find $\ p = {3\over 4} Rq \$

By the way, this confirms nicely @rude man 2011 post #6 in thread 533393 : as if the charge q was sittting in the center of mass of this top half sphere (page 5 here) at $\ z = {3\over 8} R \$ and -q from the lower half at $\ z = -{3\over 8} R \$.

Well done !

7. Dec 17, 2015

### Incand

Thanks! That's a good way to check the answer as well putting the charge at the center of mass for each half.
And thanks for the link. The second link in rude man's post is quite interesting as well, deriving the general formula for the field of a dipole. It's different from the derivation in our book and a lot more mathematical.