# Is Electric potential constant inside a conductor in all conditions?

As electric field remains the zero inside the conductor so the potential at the surface should be the same as inside, but i came with a situation which is as follows: if a spherical conductor is placed inside (concentrically) a conducting shell which has greater dimensions than that of the first conductor and a some charge is given to the smaller conductor then no work should be done as the potential remains the same however if we weld a metal conductor conducting the two spheres, then we notice that the entire charge Q must be appear on the outer sphere by Gauss's law. I'm not able to understand the two above contradicting statements. Thanks.

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– SG8
May 16 at 16:05

## 2 Answers

It's true that the charged sphere has the same potential everywhere, but it's not true that the potential is the same as any other conductor. In particular, the concentric shell around the sphere will have a potential difference with respect to the sphere, which can be eliminated by connecting the metal conductor connecting the two. The work required for the charges to flow to the outer shell will be equal to the total energy stored in the electric field between the two conductors.

• Can you please explain why there exists a potenital difference between the two conductors? Let's say a test charge is kept at the centre of the sphere we bring to it to the surface of the outer sphere, since there exists no electric field inside either of the spheres,(both conductors) W=0, leading to V=0. what's incorrect in this? Apr 16 at 22:41
• Because there is a nonzero electric field in the space between them. Draw a Gaussian surface that encloses the inner sphere but no part of the shell. Since the sphere is charged, the normal component of the electric field on that surface is nonzero. The potential difference is then the line integral of that electric field from the sphere to the shell. Apr 17 at 15:04
• Then, after we weld the metal conductor why does entire charge Q , appear on the surface of outer shell, that would make the flux outside the inner shell surface zero, leading to normal component of field to be zero as field distribution does not depend upon the area due to symmetry Apr 21 at 15:52
• All the charge $Q$ will flow to the outer shell once the conductors are welded together. The electric field between the two conductors will vanish, as you observed. The energy required to move the charges to the surface of the outer shell will be equal to the energy stored in the electric field between the shells, which is now gone. Apr 21 at 18:50
• See this: en.wikipedia.org/wiki/… Apr 21 at 18:51

As electric field remains the zero inside the conductor

That "inside" can be read in different ways. If you mean inside the bulk conductor, it is correct. If you mean in the interior portion of a hollow conductor, it may not be correct.

If you have two conducting shells, there will be zero field within the bulk metal of either shell, but there can be a field in the gap between the shells. This will be the case if the interior shell is charged.

• What's the reason behind the electric field not being zero in the gap between the shells, can we prove this mathematically ? And also, what about the electric current in steady conditions due to drifting of electrons since they experience net force Apr 17 at 0:47
• Charge the interior shell. Draw a gaussian envelope with that shell inside and the other shell outside. As the envelope surrounds a non-zero charge, and the envelope is within the gap between the shells, there must exist a field also between the shells. Apr 17 at 0:55
• Your electric current comment sounds like a different question? There's no current required at all in this question. Apr 17 at 0:57
• Sorry I meant, if we repeat the same activity replacing hollow one with solid, then there would be no charge flow from the inner to outer one right? As potenital should remain same Apr 17 at 1:04
• If there's a path across the gap, charges would move, so potential would not remain the same. Apr 17 at 1:09