Question

Two point charges \(\mathrm{q}_1\) and \(\mathrm{q}_2\) are ' \(l\) ' distance apart. If one of the charges is doubled and distance between them is halved. The magnitude of force becomes \(n\) times, where \(n\) is

• A. 1
• B. 2
• C. 8
• D. 16

Hint:

Apply Coulomb's law and compare the new force with the old force.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Coulomb's force is \(F = \dfrac{k q_1 q_2}{r^2}\).

Step 2: New Force
One charge becomes \(2q_1\) and the distance becomes \(\dfrac{l}{2}\).
\(F' = \dfrac{k(2q_1)q_2}{(l/2)^2} = \dfrac{8kq_1q_2}{l^2}\).

Step 3: Ratio
\(\dfrac{F'}{F} = 8\), so \(n = 8\).

Final Answer:
The force becomes 8 times the original. \[ \boxed{\text{(C) }8} \]