Question

The force of repulsion between two-point charges is F, when these are 1 m apart, Now the point charges are replaced by conducting spheres of radii 5 cm having the charge same as that of point charges. The distance between their centers is 1 m, then the force of repulsion will
A. increase
B. decrease
C. remain same
D. becomes 10F/9

Answer 1

Hint: Initially there are two-point charges and while calculating coulombic force between two charges we take the distance between the two charges, when the point charges are replaced by conducting spheres of some radius the charge remains the same but the distance changes.

Complete step by step answer:
From coulomb’s law \[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\], where k is a constant and r is the separation between the two charges which is given as 1m. Now, we replace these point charges by two conducting spheres, the charges remain same that is \[{{q}_{1}}\And {{q}_{2}}\] and the spheres are conducting.
Since the distance is always calculated from the centre, here the distance between the two charges increases and thus the force between them decreases.

Therefore, option B is the correct answer.

Note:
Any excess charge on a solid conductor resides entirely on the outer surface of the conductor, because of this the electric field inside the conductor is zero since the charges are free to move through the conductor. If the surface of the conductor is smooth and regular, like a sphere, the charges will push each other away until they all end up exactly the same distance from each other.