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I am trying to solve this problem:

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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.

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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.

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