Dipole moment

1. Feb 5, 2005

quasar987

Question: Consider 2 spheres of radius R and uniformly charged with density $\rho$, one positively, the other nagatively. The two spheres are partically overlapping each other and their center are separated by a distance d (< 2R). Calculate the dipole moment of this system.

What I have done so far: The dipole moment is given by

$$\vec{p}=\rho \iiint \vec{r'} \ dV'$$

I know the total dipole is the sum of the dipole of each sphere. I can calculate the dipole moment of one sphere using spherical coordinates but it becomes impossible to do so for the second sphere. Obviously rectangular coordinates is not an option.. I'm running out of ideas, what do you suggest I try?

2. Feb 5, 2005

StatusX

When the total charge is not zero, as is the case when you concentrate on a single sphere, the transformation of the dipole moment to a new origin can be derived pretty easily from the definition, and simply involves adding the total charge times the displacement vector from the old origin to the new one. Clearly the dipole moment of a single sphere is 0 if the origin is at its center, so transform it from there to wherever you like for each sphere and add them up. Since the final origin won't matter since the total charge is zero, you might as well pick it to be one of the centers to make things easier.

3. Feb 6, 2005

quasar987

I think I see what you mean... but this is precisely the resoning that led me to believe that it is impossible to get the dipole moment using that method. Let me explain where my confusion come from.

I have set the coordinate system, say S, to be centered with respect to one of the sphere (say, sphere one). With this coordinate system, the dipole moment integral is easy to evaluate using spherical coordinates. But for the other sphere, we would have to define a new coordinate system, say S*, centered on the second sphere. Then, the position vector from coordinate S is related to the position vector from coordinate S* by $\vec{r} = \vec{d} + \vec{r*}$, and the dipole moment of sphere 2 becomes

$$\vec{p}_2 = \iiint (\vec{d} + \vec{r}^*)\rho dV^* = \rho \vec{d} \iiint dV^* + \rho \iiint \vec{r}^*dV^* = \vec{d}Q_2 + \rho \iiint \vec{r}^*dV^*$$

And here you say that the second integral is 0?

I get that it's $\pi \rho R^4 \hat{r}^*$. It’s very similar to the integral for the volume of a sphere so I don’t see how I could have made a mistake here. So suppose I’ve got it right… Then what? The total dipole moment is the sum of both.. so it’s

$$\vec{p} = -\pi \rho R^4 \hat{r} + \vec{d}Q_2 + \pi \rho R^4 \hat{r}^* = \pi \rho R^4(\hat{r}^* - \hat{r}) + \vec{d}Q_2$$

Pretty ugly.

4. Feb 6, 2005

StatusX

Let me ask you this: what direction does it point (for a single sphere centered at the origin)? If you're somehow able to find a direction, rotate the problem and explain how the dipole moment could change if the setup is exactly the same. You should be able to use symmetry arguments and not have to evaluate a single integral.

Edit: I think your problem might be the r unit vector. Remember it changes with position so you can't use it inside an integral. Also, if you get it in a final answer with no specific position to base it at (remember it's undefined at the origin), you can bet the answer is wrong.

Last edited: Feb 6, 2005
5. Feb 6, 2005

quasar987

:surprised

Well, for one of the spheres centered at the origin... the dipole moment is

$$\vec{p}=\rho \iiint \vec{r} dV$$

I don't understand how you can conclude to the value of p without evaluating the integral StatusX... Could I use $\hat{r} = \hat{z}cos\theta + \hat{x}sin\theta cos\phi + \hat{y}sin\theta sin\phi$ to evaluate it?

6. Feb 6, 2005

StatusX

If you absolutely need to, you can do that integral, but the answer will just be zero. And it couldn't have been any other way.

Symmetry is a very important tool in physics because it gives you a lot of information about the answer before you even start. The guiding principle is that whatever symmetries the original problem has, any properties of the system will have them as well.

Say you have symmetry about the z axis, in that rotating the setup by any angle around it will not change the overall charge distribution. For example, a sphere or cylinder centered on the z axis has this kind of symmetry. Now, right away, there's alot you can say about the field vectors, no matter how complicated the charge distribution is. Rotating the setup by any amount around the z axis cannot affect any of the field vectors, because the charge distribution is the same. So if you know the field at one point, right away you know it at all points on a circle around the z axis through that point: just translate the vector.

But what if you're on the z axis? Can the field point in the x direction? No, because then rotating everything by 90 degrees would make it point in the y direction, but the setup is identical. The only place it can point is along the z direction, so that it isn't affected by rotation. The field at all points must have the same rotational symmetry as the setup.

Similarly, in your problem, the dipole must have the same symmetry as the sphere, which means that rotating it by any angle in any direction cannot change it. But the only vector which has this property is the zero vector. This might sound complicated, but it's really just what your intuition tells you. How could it point in any particular direction if there is no directionality in the setup? Why this direction and not another one?

7. Feb 6, 2005

quasar987

Ok, I used your logic and for the total dipole (p1 + p2), I arrived at

$$\vec{p} = -Q_2 \vec{d}$$

where Q2 is the charge of the sphere I didn't centered at the origin and d is the vector going from the origin to the center of that sphere.

It's the same dipole as that of two point charges separated by a distance d so it makes sense, right?

Last edited: Feb 6, 2005
8. Feb 6, 2005

StatusX

Yes, and as a check, verify the dipole points from negative to positive charge. Also, you can see that if you had picked the other sphere as the origin, the vector would be reversed but so would the sign of the charge, so you'd get the same answer.

9. Feb 6, 2005

quasar987

Excellent! Things are going somewhere again . Thanks for your assistance StatusX.