Two small metal spheres of equal volume are oppositely charged and are kept 0.5 m apart in vaccum. The attractive force between them is 0.144 N. But a repulsive force of 0.081 N acts between them after they are momentarily connected with a conducting wire. Find out the initial charges on the two spheres.

Explanation

Let the two initial charges be `+q_(1)` and `-q_(2)`. According to the question, the attractive force between the spheres. =0.144 N.
`therefore 1/(4pi epsilon_(0)) .(q_(1)q_(2))/(0.5)^(2) = 0.144`
or `q_(1)q_(2) = (0.144 xx 0.25)/(9 xx 10^(9)) = 4 xx 10^(-12)`
The momentary connection brings the spheres at the same potential. As the spheres have the same volume, their charges will also be the same. So, the charge on each spheres `=1/2 (q_(1)-q_(2))`.
According to the question, the repulsive force between the spheres =0.081 N
`therefore 1/(4pi epsilon_(0)) .(q_(1)-q_(2))^(2)/(4 xx (0.5)^(2)) = 0.081`
or, `(q_(1)-q_(2))^(2) + (0.081 xx 0.25 xx 4)/(9 xx 10^(9)) = 9 xx 10^(-12)`
or `q_(1)-q_(2) = 3 xx 10^(-6)`........(i)
`therefore q_(1)q_(2) = sqrt((q_(1)-q_(2))^(2) + 4q_(1)q_(2))`
`=sqrt(9 xx 10^(-12) + 4 xx (4 xx 10^(-12))`
`= 5 xx 10^(-6)`......(2)
From equations (1) and (2),
`q_(1) = 4 xx 10^(-6)` C and `q_(2) = 1 xx 10^(-6)` C
So, the initial charges on the spheres are `+4 xx 10^(-6)` C and `-10^(-6)` C