Question

The electrostatic force between two point charges at a distance of separation \(d\) is \(F\). If one of the charge is moved away by a distance \(d/2\) then the force between them is

• A. \(\frac{2}{3} F\)
• B. \(\frac{9}{4} F\)
• C. \(\frac{4}{9} F\)
• D. \(\frac{3}{2} F\)
• E. \(\sqrt{2} F\)

Hint:

If the distance is increased by a factor of \(n\), the force decreases by a factor of \(n^2\). Here, distance becomes \(3/2\) times, so force becomes \((2/3)^2 = 4/9\) times.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Coulomb's law: \(F = k \frac{q_1 q_2}{r^2}\). The force is inversely proportional to the square of the distance between the charges.

Step 2: Detailed Explanation:
Initial force: \(F = k \frac{q_1 q_2}{d^2}\). The problem states: "one of the charge is moved away by a distance \(d/2\)". This is ambiguous. It likely means the separation between the charges is increased by \(d/2\). If the initial separation is \(d\), the new separation is \(d + \frac{d}{2} = \frac{3d}{2}\). New force, \(F' = k \frac{q_1 q_2}{(3d/2)^2} = k \frac{q_1 q_2}{(9d^2/4)} = \frac{4}{9} \left( k \frac{q_1 q_2}{d^2} \right) = \frac{4}{9} F\).

Step 3: Final Answer:
The new force is \(\frac{4}{9} F\).