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

Concept: When identical conducting spheres are brought into contact, charge redistributes equally. The electrostatic force between two charges is given by Coulomb’s law: \[ F = k \frac{|q_1 q_2|}{r^2} \] Key ideas used:

Charge conservation during contact
Equal charge sharing for identical spheres
Force dependence on both charge and distance

Step 1: Initial force. Initial charges are \( q \) and \( -2q \), separated by distance \( r \). \[ F = k \frac{|q \cdot (-2q)|}{r^2} = k \frac{2q^2}{r^2} \] Since charges are opposite, the force is attractive.
Step 2: Charge after contact. Total charge: \[ q + (-2q) = -q \] Since spheres are identical, charge distributes equally: \[ \text{Charge on each sphere} = \frac{-q}{2} \]
Step 3: New separation. After separation, distance becomes \( \frac{r}{2} \).
Step 4: New force. Now both charges are \( -\frac{q}{2} \), so force is repulsive: \[ F' = k \frac{\left(\frac{q}{2}\right)^2}{\left(\frac{r}{2}\right)^2} \] \[ F' = k \frac{q^2/4}{r^2/4} = k \frac{q^2}{r^2} \]
Step 5: Compare with initial force. Initial: \[ F = k \frac{2q^2}{r^2} \] New: \[ F' = k \frac{q^2}{r^2} = \frac{F}{2} \] Hence, the force becomes half and is repulsive.