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: Understanding the Concept:
In air, the force between two point charges is given by Coulomb's law. In a medium of dielectric constant \(K\), the force is reduced by the factor \(K\).

Step 2: Key Formula or Approach:
1. In air: \(F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q^2}{d^2}\).
2. In the medium: \(F = \dfrac{1}{4\pi\epsilon_0 K}\dfrac{q^2}{D^2}\).

Step 3: Detailed Explanation:
The force is the same in both cases, so we equate the two expressions and cancel the common factors:
\[ \frac{1}{d^2} = \frac{1}{KD^2} \]
\[ KD^2 = d^2 \Rightarrow D^2 = \frac{d^2}{K} \]
\[ D = \frac{d}{\sqrt K} \]
Since \(K > 1\) for any dielectric, \(D < d\), which makes sense: the medium weakens the force, so the charges must be closer to feel the same force. Options (B) and (C) give \(D > d\), and option (D) has the wrong dimensions for a length.

Final Answer:
The distance is \(D = \dfrac{d}{\sqrt K}\), option (A). \[ \boxed{\frac{d}{\sqrt K} \text{ (A)}} \]