Two spherical conductors B and C having equal radii and carrying 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/(4piepsi_(0))(q^(2))/(r^(2))`
r, being the distance between 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`
Hence, electric force acting between B and C is
`F'=1/(4piepsi_(0))(q_(B)q_(C))/(r^(2))=1/(4piepsi_(0))((q//2)(3q//4))/(r^(2))`
`=3/8[1/(4piepsi_(0))(q^(2))/(r^(2))]=(3F)/8`