Question

Two point charges of magnitude q and 3q are located at a distance 'a' in dielectric medium of dielectric constant 2. The force applied by the larger charge 3q on dielectric is $\frac{n{q}^2}{4\pi {ϵ}_o{a}^2}.$ Here n is

Answer 1

$F_{net}$ on 3q is :
$\overrightarrow{F_{net}}=\overrightarrow{F_1}+\overrightarrow{F_2}$
where $F_1$ is Force on 3q due to q
$F_2$ is force on 3q due to dielectric

So,
$\frac{1}{4\pi {ϵ}_{o}k}\frac{3q^2}{a^2}=\frac{1}{4\pi {ϵ}_o}\frac{3q^2}{a^2}-F_2$
which gives us
$F_2=\frac{3q^2}{4\pi {ϵ}_o{a}^2}(1-\frac{1}{k})$
By putting k = 2 and comparing, we get n = 1.5