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 `r/2`. Compared to the initial force F, they will now ______.

Options

• attract with a force `F/2`.

• repel with a force `F/2`.

• repel with a force F.

• attract with a force F

Answer 1

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 `r/2`. Compared to the initial force F, they will now `bbunderline("repel with a force"  F/2)`.

Explanation:

When identical conducting spheres are brought into contact, the charge is redistributed equally. The electrostatic force between two charges is given by Coulomb’s law:

F = `k(|q_1 q_2|)/r^2`

The initial charges are q and −2q, separated by a distance r.

F = `k(|q * (-2q)|)/r^2`

= `k(2q^2)/r^2`

Since charges are opposite, the force is attractive.

Total charge:

q + (−2q) = −q

Since spheres are identical, the charge is distributed equally:

Charge on each sphere = `(-q)/2`

After separation, the distance becomes `r/2`.

Now both charges are `-q/2`, so the force is repulsive:

F' = `k((q/2)^2)/((r/2)^2)`

= `k(q^2//4)/(r^2//4)`

= `k q^2/r^2`

Comparing it  with the initial force:

F = `k(2q^2)/r^2`

F' = `kq^2/r^2`

= `F/2`

Hence, the force becomes half and is repulsive.