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. $\endgroup$ – John Rennie Nov 5 '14 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.

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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 Mar 17 '15 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. $\endgroup$ – sammy gerbil Apr 3 '17 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. $\endgroup$ – John Rennie Jan 11 '18 at 7:32
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    $\begingroup$ Has this been proven or checked by experiments? I would like to see some citation. $\endgroup$ – Anay Karnik Jan 18 '19 at 9:49

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