# Neutral Hollow Conductor in Radial External Electric Field

If a neutral hollow spherical conductor is surrounded by some external charges at some distance, it is said that charges with opposite sign will be induced on the exterior surface of the conductor and a shielding happens to the cavity inside (Faraday Cage).

Question 1: Does this mean that the conductor is no longer neutral?

Question 2: How much of these 'induced'charges can the conductor produce? What happens if it runs out of negative charges to make?

In the opposite case, where we put a charge inside the cavity, the conductor would remain neutral, since equal amounts of charge with opposite signs will be induced on the interior and exterior surfaces.

• For the second problem, I believe what I wrote here is the analytic solution. Commented Jun 19, 2023 at 4:06

Question 1: Does this mean that the conductor is no longer neutral?

No.

Question 2: How much of these 'induced' charges can the conductor produce? What happens if it runs out of negative charges to make?

The fields needed to get near to such a scenario would be go great that you would be in danger of destablising the vacuum itself. One would be hard pressed just to even create such strong fields.

We should take more time to understand what your Question 1 should be solved as. Consider one single external positive charge near a neutral spherical conductor. It does not matter if the spherical conductor is hollow or solid, the charge will only be a distribution on the surface.

What really happens is that the positive charge induces BOTH a positive AND a negative image charges on the sphere. These two induced charges have the same magnitude, i.e. cancelling out, to maintain neutrality. To have different effects, it is clear that they are at different locations. The positive charge will be at the centre of the sphere, and the negative will be closer to the external charge, along the line from the centre to the external charge, but some distance inside the sphere. The magnitude of the induced charges will be a function of the initial external charge's magnitude and the ratio of the distance between the external charge and the centre of the sphere, and the radius of the sphere. It would be at least slightly lesser in magnitude than the initial external charge.

Note that these are image charges. In reality, that actually happens is that there is a complicated distribution of charges on the surface of the sphere that gives rise to the illusion of these two image charges. That is, both positive and negative charges are distributed on the surface of the sphere. The inside is always neutral and field-less.

Actually, you have a hidden 3rd question: