Question

Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

• A. attract with a force \( \frac{F}{2} \)
• B. repel with a force \( \frac{F}{2} \)
• C. repel with a force \( F \)
• D. attract with a force \( F \)

Hint:

For identical conducting spheres: - Charges equalize after contact. - Always compare forces using Coulomb’s law carefully (charge and distance both matter).

Answer 1

The Correct Option is B

To analyze the problem, let's break it down step-by-step:

1. Initially, we have two metallic balls with charges \(q\) and \(-2q\). The force \(F\) between them when they are at a distance \(r\) is given by Coulomb's Law:

\(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{k \cdot |q \cdot (-2q)|}{r^2} = \frac{2kq^2}{r^2}\)

1. When the balls are brought into contact, they share their charge equally due to conduction. The total initial charge is \((q) + (-2q) = -q\). Thus, each ball will have a charge of \(\frac{-q}{2}\) after separation.
2. Once separated by a distance of \(\frac{r}{2}\), we calculate the new force \(F'\) using Coulomb's Law:
3. The charges are now \(\frac{-q}{2}\) on both balls. The new force \(F'\) is:

\(F' = \frac{k \cdot |\left(\frac{-q}{2}\right) \cdot \left(\frac{-q}{2}\right)|}{\left(\frac{r}{2}\right)^2}\)

Solving for \(F'\) gives:

\(F' = \frac{k \cdot \left(\frac{q^2}{4}\right)}{\frac{r^2}{4}} = \frac{k \cdot q^2}{r^2}\)

1. Comparing \(F'\) with \(F\):
2. \(F' = \frac{1}{2}F\)

Hence, the force becomes half the magnitude of the initial force \(F\), and since the charges are both negative, they repel each other.

The correct answer is: They will repel with a force \(\frac{F}{2}\).