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

Let the spherical conductors B and C have same charge q.
The electric force between them is `F = (1)/(4pi epsilon_(0))(q^(2))/(r^(2))` r, being the distance betweem them.
When third uncharged conductor A is brought in contact with B, then charge on each conductor
`q_(A) = q_(B) = (q_(A) + q_(B))/(2) = (0+q)/(2) = (q)/(2)`
When this conductor A is brought in contact with C, then charge on each conductor
`q_(A) = q_(C )= (q_(A) + q_(C ))/(2) = ((q//2) + q)/(2) = (3q)/(4)`
`F. = (1)/(4pi epsilon_(0))(q_(B) q_(C ))/(r^(2)) = (1)/(4pi epsilon_(0))((q//2) (3q//4))/(r^(2)) = (3F)/(8)`