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Induced charge on a conductive shell, potential theory

by Fernbauer
Tags: electrostatics, gauss, potential
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Fernbauer
#1
Dec31-10, 12:12 AM
P: 14
There's no electric field inside a conductor, a classic observation of electrostatics. Any field that "should" exist is compensated for by charge redistribution on the surface of the conductor. This produces classic results like shielding since in a hollow conductive shell, the field is still zero, regardless of the shell or hollow's shape.

A recent question here asked about the time it takes to establish the field.

Three unrelated questions on this common topic of electrostatics:

Does the induced surface charge change based on the presence of dielectrics, either inside or outside the shell? It seems like at least inside dielectrics must modify the charge distribution since the (induced) field inside should be affected by the dielectrics.

Is induced charge computation similar to the way point charges can be represented by charge distributions? (The classic behavior is that a shell of charge produces a field identical to a point charge outside the shell, regardless of the shell's size.) How would you calculate the "equivalent" charge density for a non-spherical shell? As a concrete example, is there some distribution of charge on the surface of the 0 centered unit cube that (to points outside the cube) behaves identically to a point charge at the origin? How would you compute that surface charge density which produces that behavior?

Last question. The mean value property of harmonic functions means you can compute the potential at a point by integrating the surrounding potentials over a closed surface or volume. The Wikipedia link shows how you'd compute the trivial case of a sphere, but how about for a cube? (ie, what weighting of points on a cube surface would integrate to the same potential at the center of the cube?) Is this weighting the same as the charge density in the previous paragraph's question?
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kloptok
#2
Jan2-11, 09:57 AM
P: 188
Now, this is not a definite answer to your questions but I think it should be important to take into account that the fields we are talking about are radially symmetric, i.e. NOT rectangularly (is that a word??) symmetric. Therefore one should not be able to extend the same line of reasoning for spherical symmetries directly to rectangular symmetries (cubes for example) when discussing electric fields.

Here are my thoughts about your questions, not really answers though I must admit:

Regarding your first question, would'nt this depend on whether the dielectric is polarized or not when putting it into the cavity? If it is not, then I guess nothing would happen with the surface charge of the shell since the field is still zero inside the cavity and the unpolarized dielectric doesn't produce a field. If it is polarized I'm not sure what would happen, but since the dielectric then produces a field of its own I would guess that this affects the charge distribution in some way. However, another possibility is that nothing happens to the surface charge on the shell and that the field inside the cavity is that due to the dielectric only (so that the E-field it is not zero in the cavity anymore).

Concerning your second question, I would say that is definitely possible to find a charge distribution for a cube that behaves identically like a point charge at the origin, but that the calculations are a bit tedious.

As to the third question, my personal intuition tells me that the weighting you're talking about should be the same as the surface charge in the preceding question but I can't prove this. (Though I must say that I find it a bit odd to try to use a weighted cube to find the mean value at the center, it feels like the point of calculating the mean value in a region as to go small, equally long distances in all spatial directions and find the values there which makes the n-dimensional sphere the natural choice of boundary surface in n dimensions.)

Feel free to discuss my thoughts and correct me if I'm wrong.

My question to you is now: What is your fascination with cubes and rectangular symmetries?
yungman
#3
Jan2-11, 12:07 PM
P: 3,898
First, I am no expert, I am learning also.

I assume there is a external E field that induce charge on the surface of the conducting shell.

For the first question, my understanding is there is no field inside the conducting shell, therefore dielectric inside do nothing to the charge. Dielectric outside the shell will polarized and generate it's or E field due to the external E field . Then the total E field experienced by the conducting shell would be the sum of both the external E and the polarizing E of the dielectric and the induced charge distribution would definitely be different from without the dielectric.

The other question, are you assume there is no external field and you have charge inside the conducting shell. You should be a little more specific so people don't have to guess.

If you are talking charge inside the shell and look at the charge distribution on the out surface of the shell. Remember the electrostatic theory stated that the surface is equal potential. It say nothing about the charge distribution. The charge distribute evenly ONLY in case of a spherical shape shell and the field behave like a point charge inside the shell. Any other shape would not result in evenly charge distribution. The shell is only equal potential on the surface no matter what shape it is. With this, I don't think the surface produce the E field like a point charge inside the shell. This is my understanding.

The third question is part of the Green's function. I think you can calculate with any shape, but good luck of setting up the equation of the surface.

As I said, I am learning too. Please comment on my post whether I am correct or not.


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