Question

The force between two electric point charges at rest in air is \( F_1 \). When the same arrangement is kept inside water, the force between them is \( F_2 \). Which of the following statement is correct?

• A. \( F_2 = 0 \)
• B. \( F_2>F_1 \)
• C. \( F_2 = F_1 \)
• D. \( F_2<F_1 \)

Hint:

In a medium with higher permittivity, the electrostatic force between two charges decreases. The dielectric constant of water is much higher than that of air, so the force in water is less than in air.

Answer 1

The Correct Option is D

The force between two electric point charges is given by Coulomb's law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] Where: - \( F \) is the electrostatic force between two charges, - \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \), - \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, - \( r \) is the distance between the charges. When the charges are placed in a medium, the force between them is modified by the relative permittivity (dielectric constant) of the medium. The modified force in the medium is given by: \[ F = \frac{1}{4 \pi \varepsilon} \cdot \frac{q_1 q_2}{r^2} \] Where: - \( \varepsilon \) is the permittivity of the medium, and \( \varepsilon_0 \) is the permittivity of free space. In air, \( \varepsilon \approx \varepsilon_0 \), but in water, \( \varepsilon_{\text{water}}>\varepsilon_0 \), meaning the permittivity in water is higher than in air. Thus, the force in water (\( F_2 \)) is given by: \[ F_2 = \frac{F_1}{\varepsilon_r} \] Where \( \varepsilon_r \) is the relative permittivity of the medium (in this case, water). Since the dielectric constant of water is greater than that of air (\( \varepsilon_{\text{water}} \approx 80 \)), the force in water will be less than the force in air. Therefore, \( F_2<F_1 \). Thus, the correct answer is Option (D): \( F_2<F_1 \).