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

Step 1: Use Coulomb's law. 
The electrostatic force between two charges is given by: \[ F = \dfrac{k q_1 q_2}{r^2} \] Step 2: Apply changes to the charges and distance. 
If both charges are halved: \( q_1' = \dfrac{q_1}{2}, \quad q_2' = \dfrac{q_2}{2} \) 
Distance is doubled: \( r' = 2r \) 
So the new force becomes: \[ F' = \dfrac{k \cdot \left(\dfrac{q_1}{2}\right) \cdot \left(\dfrac{q_2}{2}\right)}{(2r)^2} = \dfrac{k \cdot q_1 q_2}{4 \cdot 4 r^2} = \dfrac{1}{16} \cdot \dfrac{k q_1 q_2}{r^2} = \dfrac{F}{16} \]