# What is the force between two charged objects when the space between them is partially filled by a dielectric medium?

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?

• 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. Nov 5, 2014 at 16:59

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.

• @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. Mar 17, 2015 at 2:35
• 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. Apr 3, 2017 at 16:49
• @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. Jan 11, 2018 at 7:32
• Has this been proven or checked by experiments? I would like to see some citation. Jan 18, 2019 at 9:49

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.

$$\frac{\tan\alpha_2}{\tan\alpha_1}=\frac{\epsilon_2}{\epsilon_1}$$

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.) \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.