Two spherical conductors B and C having equal radii and cayying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that 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= k.(Q^(2))/(r^(2))` (fig A). Finally when a third spherical conductor come in contact alternately with `B` and `C` then removed, so charge on `B` and `C` are `Q//2` and `3Q//4` respectively (fig.B)

Now force `F' =k(((Q)/(2))((3Q)/(4)))/r^(2)=(3)/(8)F`