- #1

pc2-brazil

- 205

- 3

I was self-studying Electricity (Gauss' Law) and I have a doubt regarding the electric field near the surface of a conductor.

I know that, near the surface of an infinite plate made of a non-conductive material, the electric field can be given by:

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

However, near the surface of a conductive body, the expression becomes:

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

which is twice the expression above.

I understand how this can be explained using Gauss' law. But I'm trying to think about it only in terms of Coulomb's law.

If the conductor is an infinite thin blade, then I understand that the charges will be distributed over its entire external surface, so that the configuration will become analogous to that of two non-conductive infinite plates parallel to each other, and the electric field near the surface will be

[tex]E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2 \epsilon _0}=\frac{\sigma}{\epsilon_0}[/tex]

yielding the expression for E near the surface of a conductor. Also, between the two plates (that is, inside the conductor), E will be zero, which is consistent with Gauss' law.

But what if the conductor is a solid sphere or a solid cube, for example? When I try to think about this using only Coulomb's law, I don't see why the electric field near the surface of the conductor couldn't be given by the first expression above. As the charges are distributed over the entire external surface, if we are very near to the surface, wouldn't the configuration be analogous to that of one infinite non-conductive plate?

Thank you in advance.