What Causes the Electric Field Outside Parallel Conductors?

[mmmboh]

This isn't actually a homework question, but I have a midterm coming up and there is an example problem in the book that I don't quite get.

Basically there are two parallel conductors, each with charge Q on them, what is the electrostatic pressure on the places?

The answer is that there is no electric field between them, I imagine due to them canceling, but then on the other side of them the electric field is \sigma/\epsilon_o (well that's not the full answer obviously, but that's the part I am having a bit of trouble with)...so there are a couple things that are bothering me...

1. I know that inside a conductor the field is zero, so because of the discontinuity it is \sigma/\epsilon_o outside, but does this still apply if the plates don't have width? because then there is no inside the conductor, basically I am trying to figure out why it is \sigma/\epsilon_o outside the conductors...

And, does the electric field from the bottom plate have the same field below the top plate as it does above the top plate? Basically if you have an electric field and put some metal plate in the way, I know the charges inside the plate will move around to cancel the field inside the plate, but what about on the other side of the plate? What is the field strength? Does the induced positive charge on the right side cause the same electric field...because I would say no because it takes the positive and negative charges inside the conductor to cancel off the electric field, but I'm not sure, or does the electric field sort of pass right through?

Basically, is the field outside the metal plate \sigma/\epsilon_o because the discontinuity, or because of the influence of the other metal plate's electric field?

Thanks!

 

[fizzynoob]

>>I know that inside a conductor the field is zero
Yes if there is no current

>>so because of the discontinuity it is \sigma/e outside
Not because of continuity? Its because the material is a conductor, therefor the charges spread themselves as far away as possible from each other, creating the build up of charge on the edges of the material. That creates the field you mentioned, which you can prove using Gausses Law

>>And, does the electric field from the bottom plate have the same field below the top >>plate as it does above the top plateAnd, does the electric field from the bottom plate
The magnitudes are the same, the field direction is not.

Heres how you can solve:

\oint\vec{E}\bullet\vec{n}dA = \frac{Q}{\epsilon_{0}}

E*A = \frac{Q}{\epsilon_{0}}

\sigma = \frac{Q}{A}E = \sigma/\epsilon_{0}

 

[mmmboh]

I think I may have not been clear. I meant if you have an electric field, and you put a metal plate in the electric field, is the electric field on both sides of the plate the same, or do the charges inside the metal plate affect it?

 

[fizzynoob]

If you put a conductor in a electric field, the charge will redistribute themselves. But after this process, the E field in the conductor is not zero. There is an E-field in the conductor that is equivalent but opposite direction to the one that is passing through the conductor. That inner E-field is what moves the charges to a different distribution which momentary creates a current in the conductor. But yes if the Electric field is uniform, than the E field at the end of this process is the same magnitude on both sides

 

[mmmboh]

Ok cool, thanks! Just one thing, if you have a uniform electric field, let's say pointing to the right, and then you put in a metal plate, then negative charges will accumulate on the left side and positive charges on the right side; so wouldn't the electric field on either side of the conductor increase because of this?

 

[fizzynoob]

what it does is that it creates an electric field in the conductor that is equivalent to the E field going through it, Those E fields cancel each other and you are left with a Net E-Field in the conductor of Zero. Than on the outside you have an E-Field that is the vector sum of the E-field passing through and the E - Field created by the polarized conducting object. So yes, it does increase