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

The force between two charges in a medium of dielectric constant \(K\) is given by Coulomb's law: \[ F_{\text{medium}} = \frac{1}{4\pi \epsilon_0 K} \cdot \frac{q_1 q_2}{d^2} \] where: - \(q_1\) and \(q_2\) are the charges, - \(d\) is the separation between the charges, - \(K\) is the dielectric constant of the medium. The force between the same charges in air is: \[ F_{\text{air}} = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{d'^2} \] where \(d'\) is the equivalent separation in air. For the same electrostatic force, we equate \(F_{\text{medium}} = F_{\text{air}}\): \[ \frac{1}{4\pi \epsilon_0 K} \cdot \frac{q_1 q_2}{d^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{d'^2} \] Simplify the equation: \[ \frac{1}{K} \cdot \frac{1}{d^2} = \frac{1}{d'^2} \] This gives: \[ d'^2 = K \cdot d^2 \] Taking the square root: \[ d' = d\sqrt{K} \] Thus, the equivalent distance in air is: \[ d' = d\sqrt{K} \]