Problem 16 - A Complete Resource Book in Physics for JEE Main

Question

The ratio of the forces between two small spheres with same charges when they are in air to when they are in a medium of dielectric constant \(K\) is (A) \(1: K\) (B) \(K: 1\) (C) \(1: K^{2}\) (D) \(K^{2}: 1\)

Step-by-Step Solution

Answer

The correct answer is (A) \(1: K\).

Step 1: Coulomb's law for the force between two charges in air

Coulomb's law states that the force between two charges q1 and q2 separated by a distance r in a vacuum is given by: \(F_vacuum = \frac{k*q1*q2}{r^2}\), where k is the electrostatic constant. Since the charges are the same, we can write q1 = q2 = q, and the expression for the force between them in air will be the same as in vacuum: \(F_air = \frac{k*q^2}{r^2}\).

Step 2: Coulomb's law for forces between two charges in a medium with dielectric constant K

When the charges are placed in a medium with a dielectric constant K, the force between them is modified by the factor of 1/K. This is because the dielectric material reduces the effect of the charges on each other. The expression for the force between the charges in the medium is given by: \(F_medium = \frac{k*q^2}{K * r^2}\).

Step 3: Find the ratio of the forces in air and in the medium

To find the ratio of the forces, we will divide the force in air by the force in the medium: \(\frac{F_air}{F_medium} = \frac{\frac{k*q^2}{r^2}}{\frac{k*q^2}{K * r^2}}\).

Step 4: Simplify the ratio

We can cancel out common terms in the numerator and the denominator: \(\frac{F_air}{F_medium} = \frac{k*q^2*r^2*K}{k*q^2*r^2} = K\). So, the ratio of the forces is: \(F_{air} : F_{medium} = 1 : K\). Therefore, the correct answer is (A) \(1: K\).