For the calculation of force on one of the two equal(in magnitude, opposite in sign) point charges separated by $r$ with a dielectric slab of dielectric constant $K$ and width $d$ in between, the formula presented in many books is:-
$$F=\frac q{4\pi \epsilon_o (r-d + \sqrt{k}d)^2}$$
This means that the force between the two charges without the dielectric slab is reduced in presence of it if $k>1$. This contradicts my intuition. The charges induced on the dielectric will be due to electrostatic induction by the electric field of the two point charges and the induced charges themselves. But that means that the part of the slab towards the positive charge will have induced negative charge and the part away will have induced positive charge. Since the distance between the original charges remain unchanged, all the change in the force has to be brought about by the effect of superposition of the fields (forces) due to the induced charges on the dielectric slab. But since, considering the positive point charge, the induced negative is closer to the induced positive, there has to be a net attractive effect due to the induced dielectric charge and as a result the total force should increase rather than decrease as predicted by the formula.
Where am I going wrong? If my intuition is correct, what would the formula for finding the field be(an approximate would do)?
on a side note, would the qualitative answer (attraction increases or decreases) change if the dielectric slab is replaced by a conducting slab?
Best Answer
This appears to be a matter of wording. The force each point charge exerts on the other is indeed reduced in a sense. However, this formula does not take into account the force between each charge and the slab, which is caused by the surface charges on the slab just as you describe. This additional force can be found with the method of images.
Your question about the conducting slab is a good way to develop an intuition into this. Because the charges are opposite, if the slab is centered between them we can think of it as the limit of a dielectric material with $k$ approaching infinity. As you can see from your formula, the force between the two charges is zero with infinite $k$: no field lines will pass through the conductor to connect the two charges.
However, each charge creates an opposing surface charge on the surface of the conductor, so the charges are attracted to the conductor. If the conductor is extremely thin, the net force on each point charge will be the same as for free space. If it is thicker, the net force actually rises because the image charges created by the conductor appear to get closer to the original charges.