Question

If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$ What will be the equivalent distance between charges in air for the same electrostatic force?

• A. $k \sqrt{d}$
• B. $1 \cdot 5 d \sqrt{k}$
• C. $2 d \sqrt{k}$
• D. $d \sqrt{k}$

Answer 1

The Correct Option is D

$F=\frac{1}{\left(4\pi {\epsilon }_0\right)}\frac{q_1{q}_2}{k{d}^2}(\text{in medium)}$
$F_{\text{Air}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^{\prime 2}}$
$F=F_{Air}$
$\frac{q_1{q}_2}{4\pi {\epsilon }_{0}k{d}^2}=\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}^{{\prime }^2}}$
$d^{\prime }=d\sqrt{k}$