# Why Is the Force on Charges in Conducting Sphere Cavities Zero?

• ehrenfest
In summary, the electric field in the conducting sphere is as the charge were isolated in space, but the electric field at the charge is still zero.
ehrenfest
[SOLVED] electrostatics problem

## Homework Statement

Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral) conducting sphere of radius R. At the center of each cavity a point charge is placed--call these charges q_a and q_b.
Why is it necessarily true that the force on q_a and q_b is 0? Why is it true that the force is still zero no matter what kind of charge distribution you have outside the conductor.

## The Attempt at a Solution

You can use Gauss's Law to find the electric field in each cavity. And you find that it is as the charge were isolated in space. But that doesn't tell you the electric field at the charge...only around it, right?

*is a little rusty on the subject*

conducting sphere means that, if there is an electric field in the conducting sphere, then the electrons in the conducting sphere will move as to create an electric field to cancel the electric field in place, right?

That would answer you question, because E=0 => F=0

But q_a and q_b are in cavities, not in the meat of the conductor.

Draw a Gaussian sphere in the conductor, just outside one cavity. The E field is zero everywhere on the sphere. This can only happen if the surface charge on the inside of the cavity is spherically symmetric, i.e. constant. Then the field at the center of the spherical cavity will be zero, giving zero force on the point charge. This only happens at the center of the cavity. There would be a force on a charge at other points in the cavity.

pam said:
Draw a Gaussian sphere in the conductor, just outside one cavity. The E field is zero everywhere on the sphere. This can only happen if the surface charge on the inside of the cavity is spherically symmetric, i.e. constant. Then the field at the center of the spherical cavity will be zero, giving zero force on the point charge. This only happens at the center of the cavity. There would be a force on a charge at other points in the cavity.

I was going to say exactly the same thing.

## 1. What is an electrostatics conductor problem?

An electrostatics conductor problem is a problem that involves the distribution of charges on a conductor, such as a metal object. This can be caused by the presence of an external electric field or by the buildup of charges due to contact with other objects.

## 2. How do you solve an electrostatics conductor problem?

To solve an electrostatics conductor problem, you typically use the principles of electrostatics, such as Coulomb's law, to calculate the electric field and potential at different points on the conductor. You may also need to use boundary conditions and the properties of conductors, such as the fact that the electric field inside a conductor is zero.

## 3. What is the difference between a conductor and an insulator in relation to electrostatics?

A conductor is a material that allows charges to move freely, while an insulator is a material that does not allow charges to move easily. In electrostatics, conductors are important because charges can redistribute themselves on the surface of a conductor to create a uniform electric field inside.

## 4. Can an electrostatics conductor problem have more than one solution?

Yes, an electrostatics conductor problem can have multiple solutions. This can happen when there are multiple charge distributions or external electric fields that could give rise to the same electric field and potential on the surface of the conductor. In these cases, the solution with the lowest energy (i.e. the most stable configuration) is typically chosen.

## 5. How does the shape of a conductor affect an electrostatics problem?

The shape of a conductor can affect an electrostatics problem in several ways. For example, the curvature of a conductor can affect the distribution of charges on its surface and the strength of the electric field at different points. Additionally, the shape of the conductor can determine the boundary conditions and symmetry of the problem, which can impact the solution.

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