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: Understanding the Concept:
Coulomb's law gives \(F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{l^2}\). The force is proportional to the product of the charges and inversely proportional to the square of the distance.

Step 2: Apply the changes:
One charge doubles, so the product \(q_1q_2\) becomes \(2q_1q_2\). The distance becomes \(\frac l2\), so \(l^2\) becomes \(\frac{l^2}{4}\).
\[ F' = \frac{1}{4\pi\varepsilon_0}\frac{2q_1q_2}{l^2/4} = 8F \]

Step 3: Result:
The force becomes 8 times, so \(n = 8\).

Step 4: Why the other options are wrong.
\(n = 2\) counts only the doubled charge. \(n = 4\) counts only the halved distance. \(n = 16\) would need both charges doubled with the distance halved.

Final Answer:
The force becomes 8 times, option (C). \[ \boxed{8} \]