# Dielectric slab counter-intuitive formula?

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?

• Comment to the question (v6): Consider to add a reference for the formula to make the question more accessible. – Qmechanic Oct 22 '14 at 5:58

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.

• 1. If the formula is missing something, provide the correct formula if there is one. Actually the formula takes into account all the forces together, I could explain it but I am busy. – Rijul Gupta Jul 10 '15 at 6:57
• 2. No field lines will pass through the conductor? No not really, the conductor induces charges and even though the field lines are not inside it, the conductor does not disturb the outside field as the induced charges duplicate the original field. Again I can explain it but I am just busy. – Rijul Gupta Jul 10 '15 at 6:57
• 3. Net force increases? No it does not, the metal slab does not produce any extra electric fields! and no extra field means no extra force. Imagine a charge enclosed by a metal sphere, even though charge induces on the metal sphere equal to internal charge, only the outer charge can provide electric field and that field is equal to original field. – Rijul Gupta Jul 10 '15 at 7:07
• 1. I provided a link in my answer that tells you what is left out of the formula. – user27118 Jul 10 '15 at 18:34
• 2. The conductor does not "duplicate" the original field except in special cases (such as: a spherical conductor enclosing a point charge, or a plane conductor in a field with constant direction). This is not one of those cases. To start with, at the surface of a conductor, the electric field must be perpendicular to the conductor. The result is that a point charge near a conductor produces a completely different field than an isolated point charge. See Jackson, chapters 2 and 3, for methods to handle these problems. – user27118 Jul 10 '15 at 18:46

The correct way to understand this should be :
Imagine the 2 point charges placed $d$ distance apart, there is an electric field due to one charge which interacts with the other, now if you place the charges in a dielctric medium then near the charges opposite charges will be induced, this would produce an extra electric field in opposite direction to the original field (because the charges are opposite) the net field is therefore reduced and so is the net force.

Yes the qualitative answer for conducting slab will change, as we know conductors (uncharged) do no disturb existing fields, we can say that the force does not change in this case.

• Your second paragraph is wrong. – user27118 Jul 7 '15 at 18:51
• @user27118: please elaborate. – Rijul Gupta Jul 8 '15 at 13:58
• It is not correct that conductors do not disturb electric fields. Conductors set specific boundary conditions that change electric fields in free space. The forces on the particles will in general change, as outlined in my answer. You might be thinking of the case where we are very far from a small, neutral conductor - but that's not the case Satwik asked about. – user27118 Jul 9 '15 at 18:34
• @user27118: LOL! I didn't read your answer but if that's what you have written I should downvote it. Anyways, see this en.wikipedia.org/wiki/File:Electrostatic_induction.svg It is very clear that the outside field is not disturbed, only at the places of the metal or it's very near surroundings there is a considerable effect, nowhere else. So if satwik places a metal between two conductors there is not going to be any effect on the original force. – Rijul Gupta Jul 10 '15 at 6:52
• "You might be thinking of the case where we are very far from a small, neutral conductor - but that's not the case Satwik asked about." We are not far from the conductor, since the conductor is a large slab. In such a case the method of images is a very simple and intuitive way of understanding the resulting field. This is covered in introductory undergraduate physics courses on E&M. If it is unfamiliar to you, you need to stop responding to questions about E&M, because you will only provide disinformation. – user27118 Jul 10 '15 at 18:33

The dielectric will develop charges such that the net force in that direction stays the same I.e the along the positive plate the dielectric will develop some negative charges such that the net force remains same

• Could you give a more quantitative picture? The other answer contains a detailed discussion on the subject – Andrii Magalich Jun 20 '16 at 14:11