# Help understanding Faraday cages .

1. Aug 8, 2013

### Jimmy87

I have a basic understanding of electrostatics but can't quite get my head around how a Faraday cage works. If you take a hollow metal sphere for example and bring an external source charge towards the sphere why is there no electric field on the inside of the conductor? I know if you take a Gaussian surface of the inside then this automatically tells you there is no E-field but this gives me little intuition to what is actually happening. If the external source charge is positive then it will have electric field lines which start on the source charge and end at infinity which go radially outwards. What happens when these field lines come across the metal shell - do they go in and then back out giving zero net flux or can they physically not enter the shell? I did read somewhere that the electrons on the shell rearrange themselves to cancel the E-field of the external source charge but what exactly is meant by "cancel" - do the electrons create there own E-field or something? Many thanks in advance for any help.

2. Aug 8, 2013

### PhysicoRaj

As you bring a charge near the conducting shell, it induces charges of opposite polarity on the surface closest to it. The other charge ends up on the other side. The resulting charge configuration(as there is a separation of positive and negative charges) has it's own electric field equal and opposite to the field of the external charge. Thus the net field is zero inside the conducting shell. This is what is meant by "cancel."

As long as the external charge is absent, the equal positive and negative charges are uniformly distributed throughout the shell and hence have a zero resultant field.(not that electric fields are generated at some instance or the other. They are always present near a charge). But when the source charge interferes, the charges are separated, the E-field of each electron comes into play and the resultant non-zero E-field is observed.

3. Aug 8, 2013

### Jimmy87

Thanks for your answer, that makes more sense now. So if a positive source charge is brought close to one side of the shell, will that side of the shell have a negative net charge induced on it which then cancels the field of the source charge, is that right? Does this cancelling also mean that there is never any flux in the shell, i.e. the field lines of the source charge never penetrate the shell?

4. Aug 8, 2013

### sophiecentaur

There is a gravitational equivalent. Take a long trough of water and tilt it. Whatever the shape of the trough, the level of the water will end up horizontal. There will be no 'slopes' half way along the trough. The charges on the metal container will flow to equalise the potential everywhere so there will be no potential differences inside (or even on the surface.