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.

Answer 2

Two small identical metallic spheres carrying charges \( q \) and \( -2q \) are touched together and then separated at half the original distance. Let's work out what happens to both the sign and the magnitude of the force, and check each option against it.

1. Attract with a force \( \frac{F}{2} \): Attraction would require the two final charges to have opposite signs. But once identical conducting spheres touch, the total charge \( q + (-2q) = -q \) splits equally between them, so both spheres end up with the same sign of charge, \( -\frac{q}{2} \) each. Since both charges carry the same sign, the spheres cannot attract, this option is incorrect.
2. Repel with a force \( \frac{F}{2} \): With both spheres now negatively charged at \( \frac{q}{2} \) each, the force must indeed be repulsive, so this option gets the direction right. Checking the magnitude: originally \( F = k\dfrac{2q^2}{r^2} \). After contact, \( F' = k\dfrac{(q/2)(q/2)}{(r/2)^2} = k\dfrac{q^2/4}{r^2/4} = k\dfrac{q^2}{r^2} = \dfrac{F}{2} \). The magnitude also checks out.
3. Repel with a force \( F \): The direction (repulsion) is correct, but the magnitude is wrong, halving the charge on each sphere while also halving the distance changes the force by a net factor of \( \frac{1}{2} \), not by a factor of 1, so the force cannot stay the same as before.
4. Attract with a force \( F \): This is wrong on both counts, the final charges are like-signed (no attraction), and the magnitude does not remain unchanged.

Only the second option satisfies both the correct sign (repulsion, since the spheres now carry equal like charges) and the correct magnitude (halved, from the combined effect of reduced charge and reduced distance).

Therefore, the correct answer is they will repel with a force \( \frac{F}{2} \).