Electrostatic polarization of an axially symmetric conductor

[Diane Wilbor]

Homework Statement

A grounded Z-axis symmetric closed conductor has a single point charge at the origin within it, inducing negative charge onto its inner surface.

Given the induced charge density from the unit point charge, find the surface charge induced instead by a unit dipole at the origin, oriented in the +X direction.

Homework Equations

I'm unsure!

The Attempt at a Solution

I can write down the answer by inspection. If the surface density from the point charge has some (given!) $$\sigma(r, z, \theta) = f(r, z)$$ then the dipole induced density will influence only the $\theta$ dependence, with the obvious cosine dipole weighting, and thus the answer will be proportional to $$\sigma_{\rm dipole}(r, z, \theta)=f(r, z)\cos\theta.$$ As a check for this intuition, it does hold true for a dipole inducted charge for a conductive sphere. Spherical coordinates would be the same, but cylindrical coordinates seem more natural for this kind of problem.

The constant of proportionality is unknown, but we can integrate our (proposed) solution and find that we need to divide by the mean radius, $\int f(r,z)\ r\ dr\ dz.$

But the hard part is proving this. I first thought to expand potential in terms of a multipole expansion of the potential in cylindrical coordinate, then seeing which terms get "shifted" with a cosine weighting of $\theta$ but this doesn't go anywhere since the multipole expansion is just 0 (for the system as a whole) or a unit radial (for the conductor's contribution to the farfield expansion.)

Another interesting thought is that if the system were spherically inverted with a Kelvin transformation, we'd have the same problem but with an initial system of an axial symmetric charged capacitor and relating that surface density to the surface charge induced into the same conductor in a linear external field. Interesting, but it doesn't give any new tool to analyze the problem.

The problem specifies the conductor is closed. I suspect this is an unnecessary condition and the problem is valid for any axially symmetric geometry, closed or open, and with no need for the origin to be within the conductor's interior. Again this gives no help to the answer, but was just another observation when thinking about the problem.

I considered a simple substitution. If we replace the central charge with a +X dipole, we would still be in balance if every bit of surface charge was replaced with a proporionally scaled +X dipole as well. Those surface dipoles are not normal to the conductor. But how to transform the oriented surface dipole into an equivlent charge? The $\cos \theta$ term could come from ${p}\ \cdot\ {\hat{n}}$ but what's the justification for that?

I'm running out of tools to attack the problem. This is for self-study, so I am eager to gain more intuition about how to think about such manipulations. Thanks for any help!

 

[EricGetta]

You're wrong, because a dipole has no net charge, its got equal positive and negative charges that sum to zero. So a dipole inside of a conductor will not induce any charge density on its surface. So the answer is simply density=0.

 

[TSny]

You're wrong, because a dipole has no net charge, its got equal positive and negative charges that sum to zero. So a dipole inside of a conductor will not induce any charge density on its surface. So the answer is simply density=0.

Although a dipole has zero net charge, it does produce an electric field. The electric field of the dipole will induce charge on the inner surface of the conductor in order to make the net field inside the conducting material zero. So there will be a nonzero surface charge density on the inner surface even though the total induced charge on the surface will be zero.

 

[Diane Wilbor]

Right, the dipole induces postive charges in one direction and negative in the other. That's what my proposed $\cos \theta$ term does. But I have to prove that the math behind that hypothesis is correct.

 

[EricGetta]

The electric field of the dipole will induce charge on the inner surface of the conductor in order to make the net field inside the conducting material zero.

That makes sense and my immediate intuition was obviously wrong. But don't you mean the induced charge will make the net field outside the conductor zero? The field inside is complex.

 

[TSny]

That makes sense and my immediate intuition was obviously wrong. But don't you mean the induced charge will make the net field outside the conductor zero? The field inside is complex.

By "inside", I was referring to inside the solid conducting material. E must be zero at every point inside the material for electrostatic conditions. If the total charge inside the hollow region is zero (as for a dipole), then E will be zero also outside the conducting shell. [EDIT: Since the conductor is grounded, E would be zero outside the shell even if the net charge inside the hollow region were not zero.]

 

[Diane Wilbor]

I keep thinking about Green's functions. We don't have the greens function explicitly, but the (given) surface density does actually represent ${\partial G\over\partial N} \cdot N$. Could we convert a source Greens function to a dipole Greens function by some kind of manipulation? Changing the N dot product to an X dot product would give the right form but wrong normalization, but I can't justify why you would do that step.

I'd appreciate anyone reading this thread to post even any IDEAS for a strategy to attack this problem. I'm still stuck and I'm just unsure what approaches to research any more.

 

[EricGetta]

How about looking in a text like Jackson for a similar problem?

 

[Diane Wilbor]

Eric, great idea! And I've done exactly that, looking in Griffiths, Jackson, Smythe, and Eyges texts. I haven't spotted any problem like this one. I suspect the proof is simple and obvious, I just can't spot it.

Over the weekend I did verify the cosine solution is correct for a cylinder as well as a sphere, but that's obviously not the general case.

I'm getting a bit depressed that I can't figure this out. I don't mind doing hard work to understand it, but I just don't have any clues how to proceed. And it looks like it's stumped the viewers on this forum too!