Question

If a dielectric slab is placed between two charged particles does the force on 1 due to other change

Answer 1

A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is KK and the thickness of the dielectric is tt, then for calculating the force it behaves as if the thickness was tK−−√tK.

To see this let's take the example we know about where the dielectric fills the space between the charges:



In (a) the thickness of the dielectric is the same as the distance between the charges, rr, so the effective thickness is rK−−√rK. If we put this in the force law we get:

F=14πϵ0Q1Q2(rK−−√)2=14πϵ01KQ1Q2r2F=14πϵ0Q1Q2(rK)2=14πϵ01KQ1Q2r2

as we expect. The force is reduced by a factor of KK.

Now consider (b). To get the effective distance between the charges we have to add the distance through the air, r−tr−t, plus the effective thickness of the dielectric, tK−−√tK, so the effective distance between the charges is:

d=(r−t)+tK−−√d=(r−t)+tK

and the force is just:

F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK−−√)2F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK)2

This is how you get the force when the space between the charges is only partially filled by the dielectric.