Question

A metal sphere is suspended through a nylon thread. When another charged sphere (identical to A) is brought near to A and kept at a distance d, a force of repulsion F acts between them. Now A is brought in contact with an identical uncharged spheres C and B also brought in contact with an identical uncharged sphere D and then they are separated from each other. What will be the force between the spheres A and B when they are at a distance d/2 ?

Answer 1

Check this attachment :-)

Answer 2

F = kq^2/d^2

Explanation:

Let the first sphere having a charge contains a charge q

∵ It will prompt a similar amount -ve charge

Thus, q is the charge on the metal sphere A.

If it will get in contact with an additional uncharged sphere, the division of charges will be equal i.e. q/2 and they repel with one another.

∵ F' = $k\frac{q/2 * q/2}{\frac{d^{2} }{4} }$

F = $\frac{kq^{2} }{d^2}$

Thus, F = kq^2/d^2 with a repulsive character.

Learn more: Identical Spheres

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