Two identical conducting spheres P and S with charge Q on each, repel each other with a force 16 N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

Solution

Physics sir · 4 steps

Each touch splits the combined charge equally, so track P and S one contact at a time.

1. Before any contact. Both spheres carry Q, so F = kQ²/r² = 16 N. When two identical conductors touch, their total charge is shared equally between them.
2. R touches P. Total charge is Q + 0 = Q, so each gets Q/2. Now P has Q/2 and R carries Q/2 away.
3. R touches S. Total charge is Q/2 + Q = 3Q/2, so each gets 3Q/4. S now has 3Q/4.

   F' = k(Q/2)(3Q/4)/r² = 3/8·kQ²/r² = 3/8× 16 N

4. So. F' = 6 N, which is option (b).

The idea to keep. For identical conductors in contact, add the charges and halve the sum. With successive contacts, update the moving sphere's charge after every touch before the next one.

Answer from the NTA final answer key. Solution by PrepWiser.