Q110 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
Two equal point charges exert a force F on each other when they are placed distance 'd' apart in air. When they are placed distance 'D' apart in a medium of dielectric constant K, they exert the same force. The distance D is equal to
- Correct answerA$\dfrac{d}{\sqrt{K}}$
- B$d\sqrt{K}$
- C$d^2 K$
- D$\dfrac{K}{d^2}$
Solution
The electrostatic force between two point charges in air is given by
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2}$
When the charges are placed in a medium of dielectric constant $K$ at a distance $D$, the force is
$F' = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q^2}{D^2}$
Given that the forces are equal, $F = F'$:
$\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2} = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q^2}{D^2}$
$\dfrac{1}{d^2} = \dfrac{1}{K D^2}$
$D^2 = \dfrac{d^2}{K}$
$D = \dfrac{d}{\sqrt{K}}$
Answer: $\dfrac{d}{\sqrt{K}}$
$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2}$
When the charges are placed in a medium of dielectric constant $K$ at a distance $D$, the force is
$F' = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q^2}{D^2}$
Given that the forces are equal, $F = F'$:
$\dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{d^2} = \dfrac{1}{4\pi\epsilon_0 K} \dfrac{q^2}{D^2}$
$\dfrac{1}{d^2} = \dfrac{1}{K D^2}$
$D^2 = \dfrac{d^2}{K}$
$D = \dfrac{d}{\sqrt{K}}$
Answer: $\dfrac{d}{\sqrt{K}}$
Hence, the correct answer is option: 'A'.