# Electric field in and on a square conducting slab

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A square conducting slab with 10m sides carries a net charge of 10 micro Coulombs.

Use Gauss' law to find E inside the slab and close to its surfaces, far from the edges of the slab.

The slab is placed to the right of an infinite charged nonconducting plane with charge density of 2 X 10^-6 C/m^2 so that the faces of the slab are parallel to the plane. Find the electric field on each side of the slab far from its edges and the charge density on each face.

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##### Solution Summary

A detailed self-contained solution in which Gauss' law is explained is given.

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When the charge distribution is known, the electric field at any position is fixed by Coulomb's law. A charge of q at position x contributes to the field at position x' by an amount:

q P(x,x')/(4 pi epsilon_0 (x-x')^2),

where P(x,x') is the unit vector pointing from x to x'. So, in principle you can write down a surface or volume integral giving the electric field at any point. From this you can derive Gauss' law:

Integral over a closed surface of the normal component of the electric field equals Q/epsilon_0, where Q is the total charge contained inside the surface.

When there are conductors you have to deal with the free movement of charges. Any electric field inside a conductor would exert a force on charges causing them to move. In an equilibrium situation, the electric field inside a conductor is thus exactly zero.

On the surface of a conductor there can only be a normal component, because any parallel component would cause charges to move parallel to the surface until that component is zero.

Take a small closed surface inside the conductor. Because the electric field is zero everywhere inside, the total charge ...

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