Charge Flow in Earthed Conducting Shell with Non-Symmetric Configuration

[Brilli]

Homework Statement

Consider a thin conducting shell of radius r carrying total charge q. Two point charges q and 2q are placed on the points A and B which are at the distances 0.5r and 2r from centre of the shell respectively. If the shell is earthed how much charge flows from to the earth?

Homework Equations

The Attempt at a Solution

Generally when there is a symmetric configuration i take potentials due to all sources at the earthed surface and equate it to 0. But this is not a symmetrical problem. How to get the value?

 

[BvU]

Can't give a hint without giving it away, except perhaps: what is the answer if the two charges are placed in the exact center ? The relvant equation you use to answer that (oops, you didn't have anything at all here ... ) can be used in the non-central case as well.

 

[Brilli]

Can't give a hint without giving it away, except perhaps: what is the answer if the two charges are placed in the exact center ? The relvant equation you use to answer that (oops, you didn't have anything at all here ... ) can be used in the non-central case as well.

Will the charge distribution over the whose surface area of the sphere be uniform or non uniform?

 

[haruspex]

Will the charge distribution over the whose surface area of the sphere be uniform or non uniform?

It will be non-uniform.
Do you know about the method of images for induced charges on grounded spheres?
https://en.wikipedia.org/wiki/Method_of_image_charges#Reflection_in_a_conducting_sphere

 

[Brilli]

It will be non-uniform.
Do you know about the method of images for induced charges on grounded spheres?
https://en.wikipedia.org/wiki/Method_of_image_charges#Reflection_in_a_conducting_sphere

Thank you. I will refer to it and try it now.

 

[rude man]

Not sure, but I think the idea is to find the image charge separately for each given charge, then charge lost = original charge minus the sum of the two image charges.

 

[ehild]

Not sure, but I think the idea is to find the image charge separately for each given charge, then charge lost = original charge minus the sum of the two image charges.

Image of the charge inside the shell is irrelevant. It would influence the surface charge distribution on the inner surface of the shell and the electric field inside the shell, while earthing acts on the outer surface.

 

[BvU]

Can't give a hint without giving it away, except perhaps: what is the answer if the two charges are placed in the exact center ? The relevant equation you use to answer that (oops, you didn't have anything at all here ... ) can be used in the non-central case as well.

Need to apologize to @Brilli : I had read the exercise as if both charges were placed inside the sphere -- the 2q is not.

New proposal: same exercise, what is the answer if the 1q charge is placed in the exact center -- and we forget about the 2q charge for the moment ? So:

Consider a thin conducting shell of radius r carrying total charge q. A point charge q is placed at the centre of the shell. If the shell is earthed how much charge flows from to the earth?

From you I expect a drawing (three, actually: situation at start with 1q sphere, situation after additional 1q is placed and situation after earthing of the sphere) plus a relevant equation (or well-known theorem ... )

$\$

 

[haruspex]

Image of the charge inside the shell is irrelevant. It would influence the surface charge distribution on the inner surface of the shell and the electric field inside the shell, while earthing acts on the outer surface.

This doesn't sound right to me. If I start with a shell with no net charge but an isolated charge inside it, when I ground the shell there will be a flow of charge. You seem to be saying that the isolated charge is irrelevant.

 

[ehild]

You seem to be saying that the isolated charge is irrelevant.

No, you are wrong, I did not say that. I said that the image of the inner charge is irrelevant, The induced charge on the surface will flow away.

 

[haruspex]

No, you are wrong, I did not say that. I said that the image of the inner charge is irrelevant, The induced charge on the surface will flow away.

Ah, you mean finding the image is unnecessary, right?

 

[ehild]

Ah, you mean finding the image is unnecessary, right?

No, I said the image of the inner charge is irrelevant, answering rude man, who suggested to find both image charges. You can conclude that it is not necessary to find it. It does not mean that finding the image of the outer charge is not necessary.

 

[haruspex]

No, I said the image of the inner charge is irrelevant, answering rude man, who suggested to find both image charges. You can conclude that it is not necessary to find it. It does not mean that finding the image of the outer charge is not necessary.

Yes, I meant the image of the inner charge.
You threw me by saying that the image of the inner charge was irrelevant, which would imply that the inner charge itself was irrelevant. What you meant was that you can work with the inner charge as is - you do not need to find its image.

 

[ehild]

Yes, I meant the image of the inner charge.
You threw me by saying that the image of the inner charge was irrelevant, which would imply that the inner charge itself was irrelevant.

No, it does not imply that the inner charge itself is irrelevant.
The inner q charge induces -q charge on the inner surface, and a homogeneous charge distribution on the outer one. The induced charge on the outer surface is q. The earthing will remove the charge on the outer surface of the shell. Using the image of the inner charge you could determine the charge distribution on the inner surface, which is irrelevant for the problem.
The Method of Images uses both charges, the real one and its image to determine the electric field and the surface charge distribution on the conducting surface on the same side where the real charge is.

 

[BvU]

The inner q charge induces -q charge on the inner surface, and a homogeneous charge distribution on the outer one. The induced charge on the outer surface is q

Is what I wanted Brilli to 'discover' so he/she can find his/her way through this exercise...

 

[ehild]

Is what I wanted Brilli to 'discover' so he/she can find his/her way through this exercise...

Sorry, but I was in a debate with Haruspex. He did not seem to understand my points, so I had to explain it to him.
As for Brill, with that additional help he might be able to do some progress.