Question

in one of the MIT open lectures Walter Lewin states and proves that the electric field inside a charged conducting sphere is zero. Also if there was such a thing as a hollow planet the gravitation field inside would be zero.

My question is. Is it just that the electric / gravitational field DUE to the charge / gravity at the sphere surface is zero or is the field from other charges outside ( or other planets ) zero as well?

Answer 1

only the electric field due to the surface charge of the sphere will be zero. There will be a field at any point inside the conducting sphere if there are charges present outside it.

Answer 2

This is not true. The steady state field inside the conducting sphere will be zero no matter what charges exist outside the sphere. In fact any shape of conducting shell can be made and the field inside due to charges outside the shell will be zero. Such cages are known as Faraday Cages.

What happens is outside charges cause charges on the surface of the conducting shell to redistribute such that the field due to this new surface charge distribution and the field due to the outside charges cancel within the shell.

The Gauss' Law proof for a spherical shell I am convinced is flawed and is something that has been repeated over and over again in textbooks for decades. But there is a simple proof that is valid for any shape of conducting shell. It goes like this: Suppose there is an E field in the interior of a conducting shell of any shape. E field lines must originate and terminate on charges. The presence of these field lines in the interior of a shell with no charges present means the field lines must start and end on the interior shell's surface. But if that is the case, there is force acting on those two charges that would cause them to move through the conductor body. Therefore the shell cannot be in steady state. So, no field can exist in the interior of any enclosed conducting shell at steady state.

This same argument can be made to see no field can exist inside the shell even in the presence of exterior charges. Now, when an E field changes because charges outside the shell are moved, it does take time for the charges on the conducting shell to redistribute to cancel the interior field but for good conductors like copper, this is on the order of 10^-19 seconds -- ie virtually instant. However this finite time does mean in real life, with moving charges outside the shell, the field inside is never completely zero.

The scenario above concerns only static or slowly varying E fields. The case of propagating E fields like radio waves or whatnot is different. Now you have a rapidly changing E field that the conducting shell cannot instantaneously distribute charges to cancel. What happens instead is the finite time for the charges to move absorbs energy from the incoming waves and causes the E field to attenuate quickly. If the shell's thickness is large enough, you can reduce these waves to a negligible level before they reach the interior.

*** The case with gravity is different. With conductors, surface charges are free to move around to cancel the field within a hollow interior. With gravity, surface mass is not free to move around to cancel a gravitational field due to masses outside a hollow body. IF you could rearrange the mass in the same manner as the electric charges are rearranged then you could in principal cancel the gravitational field due to outside masses on the interior of a hollow body.

Answer 3

junkyard is right