Question

Two charges \(q_{1}\) and \(q_{2}\) are separated with distance 'd' and placed in a medium of dielectric constant 'k'. The equivalent distance between charges in air for the same electrostatic force is

• A. \(\frac{d}{\sqrt{k}}\)
• B. \(\sqrt{\frac{k}{d}}\)
• C. \(d\sqrt{k}\)
• D. \(k\sqrt{d}\)

Hint:

Equate the force in the medium with the force in air at the new distance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In a medium, \(F=\dfrac{q_1q_2}{4\pi\epsilon_0\,k\,d^{2}}\). In air at distance \(d'\), \(F=\dfrac{q_1q_2}{4\pi\epsilon_0\,d'^{2}}\).

Step 2: Equate
\(k\,d^{2}=d'^{2}\)

Step 3: Solve
\(d'=d\sqrt{k}\)

Step 4: Check
Since \(k>1\), the air distance is larger, as expected.

Final Answer:
The equivalent distance in air is d root k. \[ \boxed{\text{(C) }d\sqrt{k}} \]