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. (a) `F/8`
B. (b) `(3F)/4`
C. (c) `F/4`
D. (d) `(3F)/8`

Answer

D

Explanation

`Fprop(Q_AQ_C)/(x^2)`
x is distance between the spheres. After first operation charge on B is halved i.e `Q/2` and charge on third sphere becomes `Q/2`. Now it is touched to C, charge then equally distributes them selves to make potential same,
hence charge on C becomes `(Q+Q/2)1/2=(3Q)/(4)`.
`:.` F_(n ew)prop(Q_C^'Q_B^')/(x^2)=(((3Q)/4)(Q/2))/(x^2)=3/8(Q^2)/(x^2)`
or `F_(n ew)=3/8F`