I have a very different theoretical physics question. Suppose we place a positive charge inside a conducting shell(not at the centre and not fixed), it induces the negative charge inside the shell. Now my doubt is by symmetry, the force on the charge is zero. But as the charge induces more negative charge on the side to which it is more close than the side which is far to it(as shown in the diagram), so the side which has more negative charge density attracts it.So force is not zero. But by shell theorem as it is perfectly spherical symmetry force is zero. So which of the above is true. If any other, why?

# [Physics] Force on a charge placed in conducting shell

conductorselectrostatics

#### Related Solutions

In the case of a grounded conducting shell, it is well known that the method of images can be used to calculate how the total charge $-q$ on the inner surface is distributed. The solution is given in the wikipedia link above.

Now, you ask what happens if the the potential of the shell is fixed, but not necessarily zero. The electric field inside the conductor must still be zero. Using Gauss's law and a surface that is inside the conductor we know that there then must still be a charge $-q$ distributed over the inner surface in some way. To conserve charge there must now also be a charge $+q$ distributed over the outer surface.

We know that the electric field lines leaving the outer surface of a conductor must be perpendicular to that surface. But for a spherical outer surface, the only way this can be arranged and keep the outer surface as an equipotential is if the charge $+q$ is distributed uniformly over that surface. This spherically symmetric arrangement of charge contributes no net electric field inside the conducting shell or in its interior. i.e. the electric field *outside* the conducting shell will be exactly equivalent to that of a positive charge at the centre of the shell. Furthermore, as the contribution of this charge to the field inside the shell is zero, then it cannot alter the electric field deduced using the method of images for a grounded shell.

As the electric field at the inner surface of the shell is unchanged, then the surface distribution of charge must also be unchanged.

**Therefore nothing changes about the inner shell charge surface distribution if the shell is not grounded**.

When the point charge is not at the center of the sphere, the electric field lines will not intersect the sphere at right angles. Consequently, there is an initial component of electric field along the surface of a conductor. We know this results in a force on the charge carriers inside the conductor, and these charge carriers will re-arrange until the electric field is once again perpendicular to the conductor.

At that point you have a charge concentration (positive or negative, depending on the polarity of $q$ and whether you are looking at the point closest to $q$, or furthest from it). The polarization would result in a force on $q$ - attracted towards the point of the sphere that it's closest to, since the charge concentration there is greatest.

There is no conflict with the uniqueness theorem. The UT only states that "if the solution meets the boundary conditions, it is *the* solution". And the above describes such a solution - the boundary conditions (E perpendicular to conductor surface) are met *because* there is a redistribution of charge.

The following diagram illustrates this:

I drew the field lines inside the sphere as straight lines initially, then attempted to show how they have to bend in order to meet the conductor surface at right angles. The only way I can think of bending the field lines at B is to have a charge of the same polarity as q on the surface. There can be no net field on the dotted surface (inside the conducting shell).

So getting down to brass tacks: "can there be a region where the surface charge is positive" (assuming that $q$ is positive)?

This is a question being asked on this site - and I reproduce the diagram given (for the correct answer):

The reasoning follows from the two hints supplied:

The field inside the conductor must be zero. The charge inside the shell is off-center, and hence the charge on the inner surface of the shell will arrange itself asymmetrically to cancel the field of the large positive charge.

The field inside the conductor must be zero. To the charges on the outer surface, it is as if the inside of the conductor were completely neutral. Thus, the charges on the outer surface will feel no force other than their own mutual repulsion, and will therefore have no preferred direction.

Now if you think about this, you can see that at every point where the electric field lines from $q$ hit the surface, you need an *opposite* charge to cancel the electric field - otherwise you end up with an electric field inside the conductor. And we know that can't happen. In the diagram below, $E_1$ and $E_2$ must point in opposite directions to cancel inside the conductor.

## Best Answer

It is not perfectly symmetric so you cannot apply shell theorem here. So there is no contradiction, the positive charge will move towards the inner surface of shell due to electrostatic attraction if it is not placed at the centre of shell.