I am given two charged particles of same charge at a distance of $r$. They initially apply force $F$.

Now an infinite dielectric (of dielectric constant $4$) of width $\frac{r}{2}$ is introduced between the particles. What will be the new force?

I find this problem confusing because I have only been told about forces when its either fully dielectric or fully vaccum given by Coulomb's law. How do we get forces when only partial space is dielectric?

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    $\begingroup$ I note the close votes, but I hope my answer explains the concepts involved without just doing Sigma's homework. This seems to me to be in the spirit of the Physics SE. $\endgroup$ Commented Nov 5, 2014 at 16:59

4 Answers 4


A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$.

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, $r$, so the effective thickness is $r\sqrt{K}$. If we put this in the force law we get:

$$ F = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{(r\sqrt{K})^2} = \frac{1}{4\pi\epsilon_0}\frac{1}{K}\frac{Q_1Q_2}{r^2} $$

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

Now consider (b). To get the effective distance between the charges we have to add the distance through the air, $r - t$, plus the effective thickness of the dielectric, $t\sqrt{K}$, so the effective distance between the charges is:

$$ d = (r - t) + t\sqrt{K} $$

and the force is just:

$$ F = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{d^2} = \frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{(r - t + t\sqrt{K})^2} $$

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

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    $\begingroup$ @john_rennie but i have a question here. I see you demonstrated the fact for the special case when the dielectric fills the entire space between the charges and then extended it for the other case when it partially fills the intermediate space. But how can we prove that the dielectric still behaves as an effective vaccuum distance of t√k ? Or is it just a definition? Please explain. $\endgroup$
    – Sagnik
    Commented Mar 17, 2015 at 2:35
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    $\begingroup$ I think you have not addressed the issue in the last paragraph of the question. Your first formula applies for point charges embedded in a dielectric medium. How do you know this formula applies also for the case when the medium is partly vacuum and partly a parallel-sided slab? If this approach is invalid when the slab has thickness $r$, it will be invalid when the slab has thickness $r/2$. Your approach works for a parallel plate capacitor, but I don't see why it should work for point charges. $\endgroup$ Commented Apr 3, 2017 at 16:49
  • $\begingroup$ @Arishta: the question asks about the change in the force between two particles when a dielectric is placed between them. Placing a dielectric between two charged particles reduces the force between them because the polarisation in the dielectric reduces the effective field strength. $\endgroup$ Commented Jan 11, 2018 at 7:32
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    $\begingroup$ Has this been proven or checked by experiments? I would like to see some citation. $\endgroup$ Commented Jan 18, 2019 at 9:49
  • $\begingroup$ This is method has been disproven before, it is completely incorrect. @JohnRennie what is the proof that your method works? In the case mentioned by OP it is mentioned that the dielectric is infinite - which means it behaves as an infinite sheet with two opposite charges on the two surfaces. Thus the electric field due to polarisation charges is rather 0 outside the dielectric. You said "placing a dielectric between them reduces force between the two particles", this maybe true in some cases only. The formula you used is not for dielectric between charges, it is charges embedded in dielectric. $\endgroup$
    – Ham Lemon
    Commented Nov 17, 2023 at 6:27

Boundary conditions on a dielectric are generally very complicated. While general ideas about how polarization works and affects the overall field can be talked about, the specifics of even simple cases can be too complicated to attempt properly.

So what is the deal with electric field near dielectrics?

Well to begin with I hope you have some general idea about field $\overrightarrow{D}$ (known as electric displacement vector) which is given by: $$\overrightarrow{D}=\epsilon\epsilon_0\overrightarrow{E} + \overrightarrow{P}$$

Where P is the polarization vector which gives the general idea of dipole moment per unit volume of a dielectric. It has some special properties like its polarization density and relation with Gauss' law which make it useful to define.

Ok that's great, but why is D defined in such an arbitrary manner?

That's because combining the gauss law for P and E gives us a special property for the value

$$\oint(\epsilon\epsilon_0\overrightarrow{E}+\overrightarrow{P})\cdot \mathrm d\textbf{A}=q_\mathrm{free}$$

Where $q_\mathrm{free}$ is the free charges that have not originated due to polarization of the dielectric.Generally it is easier to deal with D and free charges than E and total charge.

How are these concepts important for the problem on hand?

$$\oint{\textbf{E}\cdot \mathrm d\textbf{l}}=0$$

$$\oint{\textbf{D}\cdot \mathrm d\textbf{A}}=q_{free}$$

These equations help us determine the boundary conditions for E and D near the boundaries of dielectrics, here vacuum helps making our calculation easier. This gives us a law reminiscent of Snell's law.


Solving the problem at hand

Now the more difficult part. Solving these cases would be complicated even with the new tools in our arsenal. We have 2 charges whose effect on the dielectric would vary the overall polarization of the slab. However, if we take an assumption that $q_1>>q_2$ this would help us solve the question without too much complication. This is because it lets us ignore the negligible polarization due to $q_2$ compared to $q_1$. (I couldn't solve the actual problem without this restriction but this may help.)

enter image description here

\begin{align} \mathrm{O'C} &= \frac{\mathrm{AC}}{\tan \alpha_1} \\ &= \frac{(r-t)\tan \alpha_1 + t\tan \alpha_2}{\tan \alpha_1} \\ &= r - t + kt \\ &= r + (k-1)t \\ \end{align}

This derivation will give you the apparent distance between the 2 charges and hence the electric field of $q_1$ on $q_2$ (Please take care to remember the assumption $q_1 >> q_2$.)

P.S. If you want to learn more about dielectric theory I suggest you check out Basic Laws of Electromagnetism Chapter 3 for a shorter summary or Purcell and Morin Chapter 10 for a more in-depth explanation. I cannot rigorously prove the equations or explain the full topic here since it is a vast thing in its entirety, so it would be a good idea to check these resources out.


Although in the situation you posted in the question, I get the same result (net force on each charge decreases) but the logic/concept used to solve it was completely different and in a different case (like if there are two opposite charges) this logic would yield a different result than when the method in the top posted answer is used (by that method, you would still be getting force decreases but I would get force increases). This is because there is a basic problem in that answer (but it is somehow verified to be correct and has many upvotes) that no one seems to catch (if I go wrong at some point, do let me know, I will be grateful).

So let's go to the basics, dielectric medium (dielectric constant k) which is infinitely long compared to the distance between two charges embedded in them. Force between two charges ($q_1$ and $q_2$) is reduced by a factor of k compared to if they were in vacuum with the same distance between them. The reason for the decrease in force is polarisation of charges (or the appearance of bound charge) in the dielectric by the individual charges. Th effective charge that applies force on $q_2$ becomes $\frac {q_1}{k}$ because of the bound negative charge appearing around $q_1$ (magnitude = $q_1 (1-\frac {1}{k})$) (for a clear explanation, refer to this)

So now, this was the force when the charges were completely inside the dielectric (which is the formula the other answer has wrongly used when the charges are not in the dielectric and rather just on their surface). But what if the dielectric is placed between the charges like in your question?

So what actually happens in a dielectric is that when an external electric field is applied, it results in the "polarization" of the medium or alignment of charges (bound to each other) in accordance with the force that acts on them due the this external electric field. Like shown in this image. In the overlap region, the distance between negative and positive charges is so small that it is practically negligible and the average value of their charge for a point outside the dielectric medium is $0$. Which means the dielectric now looks like this. You may ask why this does not work in the case of charges inside the dielectric, that is because this method of taking the average value of the charge is possible because these bound charges are very close together compared to distance between them and the outside charges. When the charges are embedded in the dielectric, we can still use this method everywhere except for very close to each of the individual free charges because of the polarised charges (which surround the free charge and cannot be neglected) (like such). This essentially becomes a net charge of initial positive charge + induced negative charge (= $\frac {initial\: positive\: charge}{dielectric\: constant\: of\: medium}$) (like this). This is the reason why we only consider the polarisation of negative charge near the positive charge and not other bound charges (like the positive polarised charge at infinity on the surface of the dielectric) because they are sufficietly far apart to be averaged or considered neutral. Here, I have marked a few pairs that we can assume to be neutralised and rest of the pairs you can see for yourself.

So the very first example I want to take is of a positive charge near the surface of a semi-infinite dielectric. If there was no vacuum then obviously the charge would experience no force because there was no external electric field. Now that a dielectric is introduced, the positive charge induces a negative charge on the surface of the dielectric (I am not concerned about the magnitude) and so the positive charge in question starts experiencing a force which means force increases.

I will be using a finite dielectric from now because I want to explain the concept/working of dielectric which would not be possible in the case of infinite dielectric (see the end of this post for the answer to your question)

Another example : let us take two positive charges (seperated by a distance 2r and of the same magnitude Q) located symmetrically at the two opposite sides of a dielectric (thickness 2x) (something like this) Initially, without the dielectric, the force on them would be $\frac {Q^2}{16\pi\epsilon_0r^2}$. First let us sketch the fields due to the given charge distribution. At exactly the midpoint of the line joining the two charges, the electric field is $0$ and originates from both the positive charges and move away from them. The field lines will thus start from the respective charge and move to infinity as they approach the midpoint (a diagram like this). When a dielectric is kept in this field, then we know that negative charge will experience a force in the oppsoite direction of the electric field and a positive charge will experience a force in the direction of the electric field. So considering all field lines and respective forces, the dipoles in the dielectric will in the allign in such a way that the net negative charge is at the surface and net positive charge is at the centre (the charges in between again cancel out). As everything is symmetric and the dielectric has to stay neutral, we can say that if the negative charge at one surface is -q, then at the other surface is also q, and so at the centre, the total positive charge is 2q. (like initially this : to finally this). Imposing the condition r ≥ x and solving for net force on any one of the Q charges, we get that the net force has decreased. So if that answer rather gives us the definitive answer as to how much the force has decreased, then why am I writing all this, you may ask.

Well, that is because of this next case, where we take one positive Q charge and another -Q charge (all symmetry, distance and stuff take same as before) (diagram). Now the electric field lines originate from Q and terminate at -Q (like this) and the force on Q due to the electric field of -Q is $\frac {Q^2}{16\pi\epsilon_0r^2}$ (attractive). So now when we sketch the dipoles in the dielectric, they come out to be like this. Now let us just find the net force on the charge +Q which is the original force + the force due to induced charges. The force due to the induced charge comes out to be in the direction of te original force (as negative induced charge is closer to positive induced charges (both have same magnitude due to symmetry and previous such arguments) and so the net induced charge force is attractive - same as the original force). Now you could apply the method given in that answer that tells you to replace the dielectric by it's equivalent (and if you do apply it here, you will notice that it would yield the similiar result as thier previous answer (force decreases)). The problem of that method is : 1) it did not consider induced charges on the surface of the dielectric and rather considered charges embedded in the dielectric (as per the formula used) and 2) even if somehow problem 1) is evaded, you cannot simply "replace" the dielectric by it's "eqiuvalent" length by finding the "equivalent" length when both charges are at the surface of it. This is because the charge distribution of the induced charges varies with distance of charges from boundary and type/shape of boundary (a common problem in electrostatics is the one dealing with these problems termed "boundary level problems", a good starting point is to give these two references a read ).

Some people do ask the interesting question, why did you add the original force when considering the net force on charge Q? Isn't the medium between it dielectric, won't that make a difference to the force? You gotta understand, the force in vacuum between $q_1$ and $q_2$ separated by distance r is $\frac {q_1q_2}{4\pi\epsilon_0r^2}$. This is because in vacuum there are no polar molecules/dipoles/free ions/electrons,etc. to impede this force on $q_1$ due to the electric field of $q_2$. In a dielectric there are dipoles that can align themselves in the presence of an external electric field and that is what impeded this vacuum force on $q_1$ due to the electric field of $q_2$. But when we consider the polarisation as we did above, with surface charges and neutral in between, we have effectively reduced the dielectric to vacuum only as the "impeders" to the vacuum force were the dipoles whose alignment and polairsation we have considered and so we can just behave as though the neutral medium in between the surface charges is just vacuum (you may call this as actually "replacing" the slab by an "equivalent").

This last case (force on two opposite charged particles after introducing dielectric between them) is actually a problem from this famous book (page number 73 in the book and 78 in the pdf, question 1) by the Russian author Sir S.S. Krotov..

As for your question, because it is a semi infinite dielectric, we can say it behaves as an infinite sheet - and because the net charge of the sheet is 0, it will not provide any electric field or net force on any of the charges (as they are outside the dielectric). As for if the dielectric is finite, I think without knowing the shape of the dielectric or its distance from both the charges, we would not be able the calculate the force as I mentioned earlier about boundary level problems. the main purpose of this answer was to make you understand the working of a dielectric and to disprove some "tricks" that can be apparently applied to solve such a complex question, because I have seen on this site that that answer was tagged multiple times on multiple questions relating to force on charges with a partial dielectric medium and so I wanted to clear this misconception.


Regarding this question I want just to write what Feynman says in volume 2 on page 10.8

""Many older books on electricity start with the “fundamental” law that the force between two charges is F = q1q2/ 4πepsilon○k r^2
(equation 10.29) a point of view which is thoroughly unsatisfactory. For one thing, it is not true in general; it is true only for a world filled with a liquid. Secondly, it depends on the fact that k is a constant, which is only approximately true for most real materials. It is much better to start with Coulomb’s law for charges in a vacuum, which is always right (for stationary charges). What does happen in a solid? This is a very difficult problem which has not been solved, because it is, in a sense, indeterminate. If you put charges inside a dielectric solid, there are many kinds of pressures and strains. You cannot deal with virtual work without including also the mechanical energy required to compress the solid, and it is a difficult matter, generally speaking, to make a unique distinction between the electrical forces and the mechanical forces due to the solid material itself. Fortunately, no one ever really needs to know the answer to the question proposed. He may sometimes want to know how much strain there is going to be in a solid, and that can be worked out. But it is much more complicated than the simple result we got for liquids"".


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