JEE Physics 2023 Question with Solution

Given: Two charges $q_1$ and $q_2$ are separated by distance $d$ in a medium of dielectric constant $K$.

Find: The equivalent separation in air for the same electrostatic force.

Using Coulomb's law in a medium:

$$F_{\text{medium}} = \frac{1}{4\pi \epsilon_0 K} \cdot \frac{q_1 q_2}{d^2}$$

In air, let the required separation be $d'$:

$$F_{\text{air}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{d'^2}$$

For the same electrostatic force:

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

So,

$$\frac{1}{4\pi \epsilon_0 K} \cdot \frac{q_1 q_2}{d^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{d'^2}$$

Simplifying,

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

Therefore,

$$d'^2 = Kd^2$$

Taking square root,

$$d' = d\sqrt{K}$$

Therefore, the equivalent distance in air is $d\sqrt{K}$.

The solution working gives this value, although the solution labels the correct option as D, which does not match the derived result. Hence the answer is taken from the source solution label.

The answer key states (1) $d\sqrt{K}$, and the full solution derivation also concludes

$$d' = d\sqrt{K}$$

However, the solution says The Correct Option is D. Since the provided instruction makes the solution the primary source, the recorded answer is D, but the mathematical derivation clearly matches option A.