Hacker News Reader

Comment by CamperBob2 | Hacker News Reader

I don't understand. Of course an electric charge inside an ideal, closed conductor cannot produce or affect a field outside the conductor, just as Feynman originally stated (and just as my hardbound 1989 printing of Feynman/Leighton/Sands still says, on page 5-9.)

Otherwise, as he goes on to point out, electrostatic shielding wouldn't be possible. What am I missing?

"I must have assumed it was grounded" doesn't make much sense either, because a charge inside such a conductor has no return path to ground. Either a current will flow outside the enclosed conductor -- in which case where's the return path? -- or it won't.

View on HN

Electrostatic shielding is actually about shielding the inside of a closed (hollow) conductor from the outside (the standard deriviation from Gauss's law).

If the conductor is non-grounded then a charge outside the conductor is fully affected by charges inside the conductor but not the position of all the charges (since the divergence of the field through a closed surface will produce a scalar field that varies with distance to the point at which we are measuring the electric potential so the only thing that matters is how far from the closed surface we are (and assuming all charges are enclosed within the closed surface we can represent them as a net charge at the center of the closed conductor).

If the conductor is grounded you are then basically changing the flux with which we can calculate the electrostatic potential at a point outside the sphere (and therefore changing how an external charge will be affected by an internal charge)

If you put a charge inside a spherical ungrounded conductor, charges will move to the inner surface of the conductor to make the electric field inside it zero. The excess charges go to the outer surface of the conductor, uniformly.

What happens differently when the conductor is grounded, though? From the point of view of a charge inside the conductor, grounding the outside just increases the available exterior surface area. The outer surface is now the whole planet, conceptually speaking. If field lines existed before, they still do, they're just spread out more.

A thought exercise: consider a van de Graaff generator. In operation, it moves charge from the 'ground' at the base of the belt to the top sphere. It can do this because there's a hole in the top sphere for the belt to pass through. If I enclose the entire generator in an even larger metal sphere, seamless and unbroken, and run it from batteries, can I draw sparks from that sphere? My intuition says 'no,' so it's interesting if that's not the case.

When the conductor is grounded, the charge on the conductor is no longer constant. It can change until the total enclosed charge in any sphere surrounding the conductor is zero.

Now take the charged sphere and connect it to ground with a wire. At the moment you connect it, what happens to the unbound charges that are on the surface of the sphere?

what happens to the unbound charges that are on the surface of the sphere?

That's what I'm asking. Are you're saying that yes, I can draw a spark from the exterior of a perfect, unbroken, hollow metal sphere containing a van de Graaff generator and some batteries, or otherwise detect that there's any electrostatic activity inside the sphere whatsoever?

(Actually I don't even see how a net charge can be produced inside such an unbroken conductor, thinking about it further. The VDG can only move charge around, it can't create a net imbalance inside the sphere. So the question becomes, what could? I have a feeling that while the corrected version of Feynman's statement is right in principle, it presumes a condition that can't actually exist in nature, like a magnetic monopole.)

> Are you're saying that yes, I can draw a spark from the exterior of a perfect, unbroken, hollow metal sphere containing a van de Graaff generator and some batteries, or otherwise detect that there's any electrostatic activity inside the sphere whatsoever?

I prefer to reason about the simplest cases but since you asked again: if the total charge inside an unbroken ungrounded conducting sphere is zero the field outside it will always be zero, no matter how many VdGs are inside it.

Reply on news.ycombinator.com