Question

Two point charges \( q_1 \) and \( q_2 \) are placed at a distance \( r \) in vacuum. The force between them is \( F \). If the distance is doubled and both charges are halved, what will be the new force?

• A. \( \dfrac{F}{8} \)
• B. \( \dfrac{F}{4} \)
• C. \( \dfrac{F}{2} \)
• D. \( \dfrac{F}{16} \)

Hint:

Coulomb’s law is inversely proportional to the square of the distance and directly proportional to the product of charges. If charges are scaled by \( a \) and distance by \( b \), the force scales by \( \dfrac{a^2}{b^2} \).

Answer 1

The Correct Option is D

The problem is addressed using Coulomb's Law, which quantifies the force \( F \) between two point charges \( q_1 \) and \( q_2 \) at a separation \( r \) as: \( F = k \cdot \frac{q_1 \cdot q_2}{r^2} \), where \( k \) is Coulomb's constant. Initially, the force is: \( F = k \cdot \frac{q_1 \cdot q_2}{r^2} \). Subsequently, the distance is doubled to \( 2r \), and both charges are halved to \( \frac{q_1}{2} \) and \( \frac{q_2}{2} \). The resulting force, \( F' \), is calculated as: \( F' = k \cdot \frac{\left( \frac{q_1}{2} \right) \cdot \left( \frac{q_2}{2} \right)}{(2r)^2} \), which simplifies to: \( F' = k \cdot \frac{\frac{q_1 \cdot q_2}{4}}{4r^2} \) and further to: \( F' = k \cdot \frac{q_1 \cdot q_2}{16r^2} \). Comparing the new force \( F' \) to the original force \( F \), we find: \( F' = \frac{1}{16} \cdot k \cdot \frac{q_1 \cdot q_2}{r^2} \), which means \( F' = \frac{1}{16} \cdot F \). Consequently, the new force is \( \frac{F}{16} \).