homework and exercises - 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?

Answer

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.