3 ms·

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.

by alecst 5y ago

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.

CamperBob2 5y ago
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.
- bkcooper 5y ago
  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.
- alecst 5y ago
  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?
  - CamperBob2 5y ago
    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.)
    - alecst 5y ago
      > 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.

by alecst 5y ago

CamperBob2 5y ago

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.

bkcooper 5y ago
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.
alecst 5y ago
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?
- CamperBob2 5y ago
  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.)
  - alecst 5y ago
    > 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.