Electric field near a point just outside a conductor

[arvindsharma]

Hello Friends,

in my textbook it is written that in electrostatics,electric field(E) due to a conductor at an external point very near to it is given by σ/ε.where 'σ' is the local surface charge density and 'ε' is epsilon.they have derived it by using a symmetrical conductor which have symmetrical charge density.my doubt is that whether this formula is valid for unsymmetrical conductors also or not?if yes,then can anyone explain me why is it so?if possible,then please give me a mathematical derivation.


Thanks

Arvind

 

[Meir Achuz]

It's true for any shape of the conductor. The proof, using Gauss's law, is quite simple. Choose a Gaussian pillbox with one face just inside the conductor, and the other face just outside.

 

[thegreenlaser]

Hello Friends,

in my textbook it is written that in electrostatics,electric field(E) due to a conductor at an external point very near to it is given by σ/ε.where 'σ' is the local surface charge density and 'ε' is epsilon.they have derived it by using a symmetrical conductor which have symmetrical charge density.my doubt is that whether this formula is valid for unsymmetrical conductors also or not?if yes,then can anyone explain me why is it so?if possible,then please give me a mathematical derivation.


Thanks

Arvind

How did they derive it? The typical derivation I've seen uses Gauss's law on a very small region of the surface--small enough that the surface looks flat and the charge density looks constant. Then it's easy to show that E = σ/ε very close to the surface because E=0 inside the conductor. You can also show that it points away from the surface using \oint \mathbf{E}\cdot d\mathbf{l} = 0.

That approach works for any surface or charge distribution as long as there aren't any discontinuities in the surface or the charge distribution.

If you haven't seen that derivation before, you should be able to find a more thorough explanation by searching for "perfect conductor boundary conditions" or something like that. It's pretty much the standard derivation, assuming you're at a high enough level to know what Gauss's law is.