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 having same radius as that of B but uncharged is brought in contact with B. then brought in contact with C and finaly 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

Intially `Q_B =Q_C =q ` and force between them F=`(1)/(4pi in _0) .(q^(2))/(r^(2))`
When another identical sphere A is kept in contact with B, it shares the charges of B and `Q_B.=Q_A. =(q)/(2) `
Now sphere A is brought in contact with C. Due to sharing of charges we have `Q_A.=Q_B= (q//2+q)/(2) =(3q)/(4) `
` therefore ` Final force between B and C,F. `= (1)/(4pi in _0) .(Q_B.Q_C.)/(r^(2)) =(1)/(4 pi in _0) ..(((q)/(2))((3q)/(4)))/(r^(2)) =(3)/(8) F `