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

Answer 1

Answer: (D)

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