Electric Field of a Conducting Slab

aurora14421
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Homework Statement



A charge Q is placed at height b abouve a plane horizontal conducting slab.

Write down the electric field, E(r), at a general point r above the slab (taking the point r=0 to be the point on the slab directly beneath the charge), and show that it satisfies the appropriate boundary conditions.

Homework Equations





The Attempt at a Solution



Ok, I know that the electric field above a plane is sigma*r/2*epsilon_zero*r, but since this slab has a thickness where there will be no electric field, does this make a difference? Can I still use Gauss' Theorem? And what does it mean by "appropriate boundary conditions"? Just that it's continuous at the surface of the slab and zero at infinity?

Just, generally, where do I start with this question?

Thanks
 
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For this type of problem one usually assumes that the thickness of the slap is negligible. In other words, one can consider the slab an equipotential plane.

Note that since this slab is a conductor an electric field will be induced on this plane. What can you say about this electric field?
 
Is this supposed to be solved by the method of image charges, or that is something completely different?

Well the charge induced at the surface of the slab will have the same magnitude as the point charge above it. But i am not sure what that tells me about the field since the geometry of the slab is different then the one of the point charge obviously

( I have problems with something very similar so thought to jump in if that's ok.. )
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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