Question

At a distance \( r \), two equal charges are kept and they exert a force \( F \) on each other. What is the force acting on each charge, if the distance between them is doubled and charges are halved?

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

Hint:

When using Coulomb's law, remember force is directly proportional to the square of charge and inversely proportional to the square of distance.

Answer 1

The Correct Option is C

The electrostatic force between two charges is given by Coulomb's law: \[ F = k \frac{q^2}{r^2} \] If the charges are halved, \( q' = \frac{q}{2} \), and the distance is doubled, \( r' = 2r \), then the new force \( F' \) is: \[ F' = k \frac{(q/2)^2}{(2r)^2} = k \frac{q^2/4}{4r^2} = \frac{1}{16} \cdot k \frac{q^2}{r^2} = \frac{F}{16} \]