What happens to the position of two attracting charges when a conductor is inserted between them?

[Rob2024]

Homework Statement

Two oppositely charged point charges hang from two pendulums and attract each other. A block of neutral conducting metal is placed in between them without touching the charges. Do the two charges move closer to each other or away from each other?

Relevant Equations

$\frac{k Qq}{r^2}$

I know this question is not constrained too well. Since it's not constrained too well. I thought I could just use a ball conductor, it induces two image charges which would increase the force experienced by either one of the hanging charges. Therefore the answer is that they will move closer to each other. Is this reasoning correct?

 

[kuruman]

Your reasoning is conceptually sound. Are you familiar with the method of image charges?

 

[Orodruin]

A big ”it depends”.

If you introduce a conducting plane exactly in the middle between the charges … nothing will happen. The conductor essentially introduces a forced equipotential surface, but the plane right in between the charges was already an equipotential plane without the sheet so the potential solution - and hence all of the fields - will be exactly the same.

The reasoning with introducing mirror charges is not generally sound. Consider the case where you introduce an infinite conducting plane in the middle betweenthe charges. You now have to solve the Poisson equation in both halves of $\mathbb R^3$ separately - ignoring the charge that is in the other half because it is not part of that solution. Each solution can be performed by extending the problem for the half-space to all of $\mathbb R^3$ and introducing the mirror charge. That mirror charge will be the same magnitude but opposite sign - effectively reproducing the original problem without the conductor.

For your setup with the sphere though (and essentially any setup that does not introduce a conductor along something that is already an equipotential) the original charges will induce a charge redistribution in the conductor such that it becomes an equipotential. This will typically lead to stronger attraction in the system.

 

[Rob2024]

Your reasoning is conceptually sound. Are you familiar with the method of image charges?

I am familiar with image charge. It's what I used in the reasoning since image charge for spherical conductor is well understood.

 

[haruspex]

I don’t think you need to assume any particular shape, but perhaps certain general aspects of it.
If there is a pair equal and opposite charges at the same point in the block such that the negative charge can move closer to the external positive charge at the same time as moving further from the external negative charge then it will do so. It will then exert a greater pull on the former and a reduced push on the latter.
The question is, are the directions of those forces such that the components in the relevant direction have the desired change? For example, consider the block having arms that extend around and close to the other sides of the pendulums.