Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor A having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is

Options

A. `F/4`
B. `3F/4`
C. `F/8`
D. `3F/8`

Answer

D

Explanation

Initially `F= (kq_1 q_2)/r^2= (kq^2)/r^2` (repulsive)
When a third conductor is connected with B, they share equal charges (As conductors are similar in size).So, charge on B is `q/2`, and on IIIrd conductor `q/2` .
When IIIrd conductor touches C, charge is equally shared,
C has a charge of `(q+q/2)/2=(3q)/4` finally
Force of repulsion is `F=(k(q/2)(3/4 q))/r^2 =(kq^2)/r^2 times 3/8=3/8 r`