Question

Two point charges \(q_1\) and \(q_2\) are '\(l\)' distance apart. If one of the charges is doubled and the distance between them is halved. The magnitude of the force becomes 'n' times, where 'n' is

• A. \(2\)
• B. \(4\)
• C. \(8\)
• D. \(16\)

Hint:

Coulomb's force varies directly with q1 q2 and inversely with r squared.

Answer 1

The Correct Option is C

Step 1: Factor by factor:
Doubling one charge multiplies the force by 2.

Step 2: Distance factor:
Halving the distance multiplies the force by $\left(\frac{1}{1/2}\right)^2 = 4$.

Step 3: Combine:
$2\times4 = 8$, so $n = 8$.

Final Answer:
n equals 8, option (C). \[ \boxed{8} \]