Dear Arpan,

The physical idea behind your question, as I have understood it, is the following:

1. The conducting plane is assumed to have a gas of free electrons (negative electric charge), thus if the electric charge above the plane is negative means that the density of charge seen by the charge on the plane is positive and therefore attractive force must appear between the charge-conductor plane.

2. If the electric charge above the conducting plane is positive, then the interaction force between charge-conducting plane must be non attractive.

3. If you do not assume infinite long the conducting plane, then you needed to take into account the boundary conditions. Another consideration that you do not ask but it is important is how many free electrons you have in the conductor.

I hope that this can help you to take the physical point of view of this question.

Thank You

Hello Arpan:

Problems in electrostatics such as this one often appear

mysterious but are quantitatively well explained using Maxwell's

equations. I think here you are looking for qualitative

answers: Why is the charge attracted? Why should the plane

need to be infinitely extended?

I would like to approach this problem using a basic tenet

of physics: energy conservation. But first let's clarify

few things.

- A conducting surface forms an equipotential surface (for

the elctrostatic potential).

- The potential difference between two conducting surfaces

is proportional to their charge difference.

- The electric field is conservative, i.e., the work needed

to transport a charge from one location to another

depends only on the potential difference between those

locations but not on the particular path taken to transport

the charge.

Whereas the first item is a definition the other two follow

directly from Maxwell's equations.

Now I want to understand how much work is needed to transfer

a particular charge from one conductor to another.

Transporting a tiny charge dQ requires the work V(Q)dQ, where

V(Q) is the potential difference between the conductors and

Q is the charge that was already transported (we start here

with zero of course).

Now we know that the potential difference is proportional

to the charge V(Q)=Q/C where C is called the capacitance.

The capacitance is independent of the charge but depends on

the geometry, i.e., shape and relative position, of the

conductors.

The total work needed to transport the charge Q from one

conductor to another is then

W = \int_0^Q V(q)dq = 1/2C Q2 .

This is also the energy stored in the electric field

surrounding the conductors.

This result immediately shows that the sign of the charge

is irrelevant. The work needed is the same for positive

or negative charge.

The force acting on the center of masses (cms) of the

conductors is then given by the negative gradient of the

potential energy of the two conductors. Now finally we use

energy conservation to see that the potential energy is equal

to the work that was required to transport the charge.

So far we were able to distill your question to the

following point: How does the capacitance C behave

with distance between the cms of the conductors?

This behavior is not expected to be fundamentally different

for infinite or finite planes. Therefore "infinite" is not

significant.

If the capacitance increases/decreases with distance the conductors

experience repulsion/attraction. But which one is it?

The answer is of course quantitatively given by Image Theory

which itself is derived from Maxwell's equations.

But may be we can develop a intuitive solution by having

another look at the definition of the capacitance C = Q/V.

If we keep the charge constant but move the conductors

apart will the potential between them increase or decrease?

If we keep the potential constant but move the conductors

apart will the charge increase or decrease?

For your geometry, my intuition tells me that conductors

that are far apart can only take on very little charge at

a given potential compared to conductors that are quite close.

This implies an attractive force between the conductors.

If you don't share this feeling that's ok because there's

always Image Theory to get an exact answer.