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

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Solution

Q.
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: [2025]

1 $\frac{F}{2}$

2 $\frac{3 F}{8}$

3 $\frac{3 F}{5}$

4 $\frac{2 F}{3}$

Ans.
(2)

According to Coulomb's law, $F = \frac{k q^{2}}{R^{2}}$

By touching an uncharged sphere C with charged sphere A, charge on each sphere A and C becomes $\frac{q}{2}$ .

Now, when charged sphere C is kept in contact with sphere B,
then total charge is, $\frac{q}{2} + \frac{q}{1} = \frac{3 q}{2}$

Then,

Charge on B = $\frac{3 q}{4}$

Charge on C = $\frac{3 q}{4}$

Now, charge on A is $\frac{q}{2}$ and charge on B is $\frac{3 q}{4}$ .

So, $F' = \frac{k}{R^{2}} (\frac{q}{2}) (\frac{3 q}{4}) = k \frac{3}{8} \frac{q^{2}}{R^{2}} or F' = \frac{3}{8} F$

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