Question

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

• A. \(\frac{d}{\sqrt{K}}\)
• B. \(d\sqrt{K}\)
• C. \(d^2K\)
• D. \(\frac{K}{d^2}\)

Hint:

A dielectric of constant K divides the force by K, so the distance must shrink by the square root of K to keep F the same.

Answer 1

The Correct Option is A

Step 1: Compare the two forces by a ratio:
$\dfrac{F_{\text{medium}}}{F_{\text{air}}} = \dfrac{d^2}{KD^2}$ for the same charges.

Step 2: Set the ratio to 1:
Equal forces mean $d^2 = KD^2$, so $D = d/\sqrt K$.

Step 3: Dimension check:
Distance must come out as a length. Only option (A) and (B) contain $d$ to the first power, and (B) would make the distance larger in a medium that weakens the force. Option (C) has $d^2$ and (D) has $1/d^2$, so neither is a length.

Final Answer:
Option (A). \[ \boxed{\frac{d}{\sqrt K} \text{ (A)}} \]