If two charges q₁ and q₂ 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?

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k $\sqrt{d}$

1.5d $\sqrt{k}$

2d $\sqrt{k}$

d $\sqrt{k}$

answer is D.

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Detailed Solution

F_air = F_Dielectric
$\frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{d_{a}^{2}} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{{kd}^{2}} \Rightarrow d_{a} = \sqrt{k} d$

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