# When do we consider the dielectric constant while using Coulomb's law?

I am trying to solve this problem:

The answer given is an option (D).

I had learned that whenever there exists a medium whose dielectric constant is $$K$$ then the Coulomb force becomes $$\frac{F}{K}$$ where $$F$$ is the force if the medium were a vacuum. So why shouldn't the intervening sphere affect the force between the two charges?

I had come up with a thought that if there is a continuous medium between two point charges only then would the force $$F$$ be divided by $$K$$. But I couldn't find any strong proofs for what I came up with and I feel in all probability that it isn't correct.

The Coulomb force that acts on the charge $$q_2$$ is equal to the electric field at the position of $$q_2$$ multiplied with the charge $$q_2$$. This becomes $$F = Eq_2 = \frac{1}{4\pi \varepsilon_0 k} \frac{q_1}{ r^2} q_2$$ with $$E$$ the electric field, $$\varepsilon_0$$ the vacuum permittivity, $$k$$ the dielectric constant of the material and $$r$$ the distance between the two charges. The dielectric constant $$k$$ takes into account the influence of the material onto the electric field (and thus also onto the Coulomb force).
So, if the charge was placed inside a material with dielectric constant $$k$$, the electric field would have been reduced by a factor $$k$$ and thus also the Coulomb force would have been reduced by a factor $$k$$. If the charge is placed into a vacuum, there is no influence on the electric field and thus also no influence on the Coulomb force (this corresponds to $$k=1$$). This second case is the case of your question and thus $$F_1=F_2$$.
Only the electric field at the position of $$q_2$$ is of importance. Therefore, the material in between the two charges has no influence on the force because it does not affect the electric field outside.