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?

Electrostatics PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q53 with Solution

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Correct answer: A

For two charges $q_1$ and $q_2$ separated by distance $d$ in a medium of dielectric constant $K$ , Coulomb's law gives

F_{\text{medium}}=\frac{1}{4\pi\varepsilon_0 K}\frac{q_1 q_2}{d^2}

Suppose the equivalent separation in air is $r$ . Then the force in air is

F_{\text{air}}=\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}

We want the force to be the same in both cases:

F_{\text{air}} = F_{\text{medium}}

So,

\frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = \frac{1}{4\pi\varepsilon_0 K}\frac{q_1 q_2}{d^2}

Cancel common terms $\dfrac{1}{4\pi\varepsilon_0}$ and $q_1 q_2$ :

\frac{1}{r^2} = \frac{1}{K d^2}

Thus,

r^2 = K d^2

r = d\sqrt{K}

Hence, the correct option is A.