Physics Question #2251

Paul, a 16 year old male from Romania asks on September 20, 2004,

What i don't understand is why the intensity of the electric field is 0 inside a charged (empty on the inside) metal sphere. I know that the electrical charge is spread uniformly across its outer surface ( what exactly does that have to do with it?) and that anywhere outside E=k*q*q/(r*r), but why, for example, does E(a)=E(b)=0? (a and b are two points from inside the sphere) Shouldn't the local electric charge closest to a create a very high intensity field (~/(r^2); r->0 => E->oo) in a?; while in point b (I considered point b in the middle of the sphere), because of the simetry, it's clear that E=0, thus E(a)>E(b)=0...?? I'm confused; why doesn't Coloumb's law apply here as well? If we were to consider, let's say, 50 electrified particles, placed in the form of a sphere, would it be the same as if they were on an empty metal sphere with the same radius? I guess not, but why? If you could please be somewhat explicit on the matter, and give me some detalis. I know it has to do with the sphere being a metal conductor (electrons moving freely inside it ), but my teacher simply says: "That's how things are." So how can i understand? I know "that's how things are", but "why"? And one more thing, define interior! If we were to consider a semisphere how would things be then? Or a cube? Or something like "Faraday's cage"?

viewed 14890 times

The answer

John Jones answered on September 23, 2004

The electric field at a point is just the force that a unit electrical charge at that point would experience.

So if we have an isolated conductor, any charges inside the conductor are free to move, and they will move until they don't experience any more force. At that point, the electric field inside the conductor is, by definition, zero. Since like charges repel each other, the charges will try to get as far from each other as possible, which is what pushes them to the outside of the conductor. This is why the charged sphere doesn't behave the same way as a spherical arrangement of charged particles -- the particles would at once start moving away from each other, due to their mutual repulsion, and go on moving apart forever.

This shows that the field inside the metal itself is zero, but we have not yet shown that the field inside an empty cavity enclosed by the metal is zero.

It is essential that the cavity be empty; if you had a charged object inside the cavity, supported by an insulating stalk, then there would be an electric field around that object. So we'll insist that the cavity really is empty.

Suppose there were a non-zero electrical field inside the cavity. Then we could imagine taking a charged particle somewhere on the interior surface of the sphere, at a point 'A', say, and move it along a path through the field to some other point on the interior surface, 'B', say. The particle would experience a force as it moved along the path, so we would either have to do work on it to make it move along the path, or we could extract work from it as it moved along the path. Let's suppose we extract work from it -- we can change the charge on our imaginary particle, if necessary, to make this true.

We can now get the particle back from B to A by moving it through the conductor, where we know there isn't any field. So the trip from B to A doesn't involve any work. Now we have a loop, A to B to A again, and each time we go round the loop, we can extract work. So, if there were a field inside the cavity, we could use it to get energy from nothing. But this violates the principle of conservation of energy, so our initial assumption, that there was a non-zero field inside the cavity, must have been wrong.

This argument works for ANY closed conducting shell -- it doesn't have to be spherical. It does have to be complete, though; if we use a wire mesh, as in a Faraday cage, the field inside the mesh may be non-zero at a distance from the mesh up to two or three times the spacing of the wires making up the mesh.

Add to or comment on this answer using the form below.

Note: All submissions are moderated prior to posting.

If you found this answer useful, please consider making a small donation to