# Induced surface charge on a dielectric

1. May 18, 2013

### camcors

1. The problem statement, all variables and given/known data

A point charge Q is placed at a perpendicular distance a from an infinite planar LIH (linear, isotropic, homogeneous) dielectric of relative permittivity ε. By considering the normal component of the electric field at an arbitrary point just inside the dielectric to be the superposition of the normal components of the electric fields produced by the point charge Q and the induced surface charge
σ on the dielectric, show that

σ=- (ε-1)Qa /2∏(ε+1)r^3

2. Relevant equations

Normal component due the point charge will be:

E=Qa/(4∏ε0εr^3)

Induced surface charge due to the electric field from Q will be:

σ=P.n (P is the polarisation)
σ=(ε-1)E

The induced surface charge produced its own electric field which is obtained from gauss' law and is just the surface charge divided by ε0.

3. The attempt at a solution

So the difference between these two fields gives the resultant field at the point just inside the dielectric. But I'm uncertain as to why the questions then wants another value for surface charge density, will this change because of the resultant field. I thought this was induced anyway such that it already accounts for the field it produced and that of the point charge. I'm therefore confused as to how to proceed.

2. May 18, 2013

### camcors

So yes just to clarify, I think the two fields I've calculated are correct. However, I don't really understand why the question is asking for the induced surface charge in this way when you already know it from the dotting the polarisation with the unit normal. Am I confusing some of the concepts here?

3. May 19, 2013

### TSny

Hello camcors and welcome to PF!

Shouldn't the second equation above be σ= εo(ε-1)En where En is the normal component of the total electric field just inside the surface of the dieletric?

I think the idea of the problem is to find σ by using this equation and writing En as the superposition of two electric fields: (1) the normal component of the field of the point charge and (2) the field due to the surface charge density σ.

Then you will have an equation with σ on both sides and you can solve for σ.

Yes, but the field produced by the surface charge σ is not quite σ/εo. There should be a factor of 2 somewhere in there. (You might be thinking of the field at the surface of a conductor with charge density σ.) If you have a flat, infinite sheet of uniform charge density σ and use Gauss' law to find E on each side of the sheet, what do you get?

4. May 25, 2013

### camcors

Yes that's what I meant slight typo! However, the way I defined E was already as the normal component. Thanks for your help I've solved this now, you're right it was a case of solving for sigma by including it in both terms.