# Homework Help: Why is the electric field inside a hollow conductor zero?

1. Mar 10, 2012

### iampaul

1. The problem statement, all variables and given/known data

The electric field is zero within the conductor because the charges are all at rest in an electrostatic situation. But the electric field inside a cavity within the conductor is not necessarily zero because it isn't part of the conductor, as my book says. Then i encountered a topic about electrostatic shielding and it says that the electric field is zero inside an empty cavity with no charge.

2. Relevant equations
∫EdA= Qenclo

3. The attempt at a solution
I tried using spherical gaussian surfaces inside the cavity: and EA=Qencl,
Qencl=0,so E=0. I can make gaussian surfaces like this anywhere inside the cavity so E must be zero everywhere. But i think my solution is wrong because it means that Electric field is always zero. Let's say that we have a gaussian surface enclosing zero charge. Then even if there is an electric field outside, the electric field must be zero on the gaussian surface. Help please.

Last edited: Mar 10, 2012
2. Mar 11, 2012

### GothFraex

The electric field within the cavity will be zero, as long as there are no charges inside. Your approach using Gauss' Law is correct. If there are no charges within the Gaussian surface, then the electric field is zero. I imagine that your book was stating that the cavity isn't part of the conductor to emphasize how remarkable this result is.

3. Mar 11, 2012

### rude man

Yes. Run a Gaussian surface around the cavity with the surface totally enclosed by metalization. You know the E field is zero within the metal everywhere. Note that this is true irrespective of the presence of charge within the cavity. This tells you that there can be no net charge within the closed surface because ε∫E*ds = qnet so qnet = 0.

The only way to get net zero charge thruout the Gaussian surface is to either have zero charge inside the cavity, or if there is charge, the induced opposite-polarity charge density on the inner surface of the cavity will distribute so as to make the above integral still true. In the absence of a finite charge within the cavity, no such induced charges can be generated. And so in that case there can be no E field within the cavity.