Question

When two charges are kept in air medium, at certain distance d apart, the force between them is F. When they arc kept in a dielectric medium at the same distance of separation, the force between them becomes F/2. Thef, the dielectric constant of the medium is

• A. 5
• B. 2
• C. 4
• D. 3
• E. 8

Answer 1

The Correct Option is B

Dielectric Constant and Electrostatic Force 

We have two charges separated by a distance \(d\). In air, the force between them is \(F\). When placed in a dielectric medium at the same distance, the force becomes \(F/2\). We need to find the dielectric constant of the medium.

Step 1: Electrostatic Force in Air (Vacuum)

The electrostatic force between two charges \(q_1\) and \(q_2\) separated by a distance \(d\) in a vacuum (or air, which is a good approximation) is given by Coulomb's Law:

\(F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{d^2}\)

Where:

• \(F\) is the electrostatic force in air.
• \(\epsilon_0\) is the permittivity of free space.
• \(q_1\) and \(q_2\) are the magnitudes of the charges.
• \(d\) is the distance between the charges.

Step 2: Electrostatic Force in a Dielectric Medium

When the charges are placed in a dielectric medium, the electrostatic force is reduced by a factor of the dielectric constant \(K\). Let \(F'\) be the force in the dielectric medium:

\(F' = \frac{1}{4\pi\epsilon_0 K} \frac{|q_1 q_2|}{d^2}\)

Where \(K\) is the dielectric constant (also known as the relative permittivity \(\epsilon_r\)).

Step 3: Relate the Forces

We are given that the force in the dielectric medium is \(F/2\):

\(F' = \frac{F}{2}\)

Step 4: Solve for the Dielectric Constant

Substitute the expressions for \(F\) and \(F'\):

\(\frac{1}{4\pi\epsilon_0 K} \frac{|q_1 q_2|}{d^2} = \frac{1}{2} \left( \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{d^2} \right)\)

Notice that the terms \(\frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{d^2}\) appear on both sides, so we can cancel them out:

\(\frac{1}{K} = \frac{1}{2}\)

Solving for \(K\), we get:

\(K = 2\)

Conclusion

The dielectric constant of the medium is 2.

Answer 2

The force between two charges in a medium is given by Coulomb’s law: \[ F_{\text{medium}} = \frac{F_{\text{air}}}{K} \] Where:
\( F_{\text{air}} = F \) is the force in air
\( F_{\text{medium}} = \frac{F}{2} \) is the force in dielectric medium
\( K \) is the dielectric constant of the medium Now substitute the values: \[ \frac{F}{2} = \frac{F}{K} \Rightarrow K = 2 \]