Physics - Electrostatics Question with Solution | TestHub

Force between charges in a medium of dielectric constant $K$ is given by $F=\frac{1}{4\pi {ϵ}_{0}K}\frac{q_1{q}_2}{d^2}$.

Force between charges in air at an equivalent distance $d^{\prime }$ is given by $F^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{d^{\prime 2}}$.

For the same electrostatic force, $F=F^{\prime }$.

$\frac{1}{4\pi {ϵ}_{0}K}\frac{q_1{q}_2}{d^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{d^{\prime 2}}$

$\frac{1}{K{d}^2}=\frac{1}{d^{\prime 2}}$

$d^{\prime 2}=K{d}^2$

$d^{\prime }=d\sqrt{K}=d[K{]}^{1/2}$

Answer: $d[k{]}^{1/2}$