Q105 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
Two point charges placed in air separated by distance $R$ exert a force $F$ on each other. Now the space between those charges is filled with dielectric constant $K$. At distance $R_1$ between the charges, the same force is exerted as $F$. Then
- A$R_1 = \sqrt{K}R$
- B$R_1 = R$
- C$R_1 = \dfrac{R}{K}$
- Correct answerD$R_1 = \dfrac{R}{\sqrt{K}}$
Solution
Force between two point charges in air separated by distance $R$ is given by:
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{R^2}$
When the space is filled with a dielectric of constant $K$, the force at distance $R_1$ is:
$F' = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q_1 q_2}{R_1^2}$
Given that the force remains the same, $F = F'$:
$\dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{R^2} = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q_1 q_2}{R_1^2}$
$\dfrac{1}{R^2} = \dfrac{1}{K R_1^2}$
$R_1^2 = \dfrac{R^2}{K}$
$R_1 = \dfrac{R}{\sqrt{K}}$
Answer: $R_1 = \dfrac{R}{\sqrt{K}}$
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{R^2}$
When the space is filled with a dielectric of constant $K$, the force at distance $R_1$ is:
$F' = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q_1 q_2}{R_1^2}$
Given that the force remains the same, $F = F'$:
$\dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{R^2} = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q_1 q_2}{R_1^2}$
$\dfrac{1}{R^2} = \dfrac{1}{K R_1^2}$
$R_1^2 = \dfrac{R^2}{K}$
$R_1 = \dfrac{R}{\sqrt{K}}$
Answer: $R_1 = \dfrac{R}{\sqrt{K}}$
Hence, the correct answer is option: 'D'.