Question

If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$ What will be the equivalent distance between charges in air for the same electrostatic force?

• A. $k \sqrt{d}$
• B. $1 \cdot 5 d \sqrt{k}$
• C. $2 d \sqrt{k}$
• D. $d \sqrt{k}$

Answer 1

The Correct Option is D

To find the equivalent distance between charges in air for the same electrostatic force, we need to understand the impact of the dielectric medium on the force between two charges.

The formula for the electrostatic force between two charges in a medium with a dielectric constant \( K \) is given by:

\(F_{\text{medium}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{K d^2}\)

Where:

• \( F_{\text{medium}} \) is the force between charges in the medium.
• \( \epsilon_0 \) is the permittivity of free space.
• \( q_1 \) and \( q_2 \) are the charges.
• \( d \) is the distance between the charges in the medium.
• \( K \) is the dielectric constant of the medium.

In air (or vacuum), the force between the same two charges at distance \( d_{\text{air}} \) is:

\(F_{\text{air}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{d_{\text{air}}^2}\)

For the forces to be equal, we set:

\(\frac{q_1 q_2}{K d^2} = \frac{q_1 q_2}{d_{\text{air}}^2}\)

Simplifying gives us:

\(d_{\text{air}}^2 = K d^2\)

\(d_{\text{air}} = d \sqrt{K}\)

Thus, the equivalent distance between the charges in air for the same electrostatic force is \(d \sqrt{K}\).

Therefore, the correct option is \(d \sqrt{K}\).