Sankalp NEET Full Test-3 Question-28 Solution

Question: 28. Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

(1) \frac{3 F}{5}

(2) \frac{2 F}{3}

(3) \frac{F}{2}

(4) \frac{3 F}{8}

Answer: Option (4)

Explanation:

Initially, each of the two identical spheres A and B has charge q. The force of repulsion between them is F.

Using Coulomb’s law, this can be written as

F = k \frac{q^{2}}{r^{2}},

where k is Coulomb’s constant and

r is the distance between their centres.

When the third identical uncharged conducting sphere is brought in contact

with sphere A, total charge to be shared is q

(because A has charge q and the third sphere has zero charge).

Since the two spheres are identical, the charge divides equally:

\text{Charge on } A = \frac{q}{2} \text{Charge on third sphere} = \frac{q}{2}

Now, sphere B is still at charge q.

The third sphere with charge \frac{q}{2}

is then brought in contact with B.

The total charge to be shared between them is:

\frac{q}{2} + q = \frac{3q}{2}

Again, since they are identical conducting spheres, this charge is shared equally:

\text{Charge on } B = \frac{\frac{3q}{2}}{2} = \frac{3q}{4} \text{Charge on third sphere} = \frac{3q}{4}

After this, the third sphere is removed. So the final charges on the two original spheres are:

\text{Final charge on } A = \frac{q}{2} \text{Final charge on } B = \frac{3q}{4}

The new force of repulsion between A and B is given by Coulomb’s law:

F_{\text{new}} = k \frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^{2}} = k \frac{\frac{3q^{2}}{8}}{r^{2}} = \frac{3}{8} k \frac{q^{2}}{r^{2}}

But F = k \frac{q^{2}}{r^{2}}, so:

F_{\text{new}} = \frac{3}{8} F

Therefore, the new force of repulsion between spheres

A and B is

\frac{3 F}{8},

which corresponds to Option (4).