CBSE Class 12-science Answered
Let the two charges be q1 and -q2
in the first case,
F = -kq1q2 / r2
here,
r = 1m
so,
F = -kq1q2 /12
or
F = -kq1q2 .................(1)
and
when the spheres are connected by wire the charge transfer takes place. It is such that the charge now is equally distributed over the two spheres
so,
charge on each sphere will be = (q1 + q2)/2
thus, the coulombic force will be
F' = k((q1 + q2)/2).((q1 + q2)/2) / 12
or
F' = F/3 = k(q1 + q2)/2)2
or
F/3 = k(q1 + q2)/2)2
or
F = 3k((q1 + q2)/2)2..............................(2)
now, by equation (1) and (2), we get
-kq1q2 = 3k((q1 + q2)/2)2
or by solving further we get
3[(q1 + q2)/2]2 = - q1q2
or
3[q1 2 + q2 2 + 2 q1q2] / 4 = -q1q2
or
3[q1 2 + q2 2 + 2 q1q2 ] = -4q1q2
or
3q1 2 + 3q2 2 + 10q1q2 ] = 0
now, by diving both sides by q2 2, we get
3(q1/q2)2 + 10(q1/q2) + 3 = 0
let q1/q2 = x
so, we have
3x2 + 10x + 3 = 0
which is a form of a quadratic equation, the roots of which are
x = -0.33 and -3
in magnitude
|x| = 0.33 or 3
thus, the ratio intial of charges will be
q1/q2 = x = 0.33
or
q1/q2 = x = 3
Therefore the ratio is 3:1