Question

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. \(R_1 = \frac{R}{K}\)
• D. \(R_1 = \frac{R}{\sqrt{K}}\)

Hint:

A dielectric reduces the force by K, so the distance must shrink by root K to restore the force.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In air, \(F=\dfrac{kq_1q_2}{R^2}\). In a medium of dielectric constant \(K\), the force is reduced by the factor \(K\).

Step 2: Force in the medium
\[ F'=\frac{kq_1q_2}{KR_1^2} \]

Step 3: Set equal to the force in air
\[ \frac{kq_1q_2}{KR_1^2}=\frac{kq_1q_2}{R^2}\Rightarrow KR_1^2=R^2 \]
\[ R_1=\frac{R}{\sqrt K} \]

Step 4: Check the options
Since \(K>1\), the charges must be brought nearer. That rules out \(\sqrt KR\) and \(R\). Option (C) would be right only if the force were proportional to \(1/R\). The answer is option (D).

Final Answer:
The force in the dielectric is k q1 q2 over K R1 squared, so R1 must equal R over root K, option (D). \[ \boxed{R_1=\frac{R}{\sqrt{K}}} \]