The centres of 2 identical small conducting spheres are 1m apart. They carry charges of opposite kind and attract each other with a force F. When they are connected by a conducting thin wire they repel each other with a force F/3. What is the ratio off magnitude of charges carried by the spheres initially?
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[q12 + q22 + 2 q1q2] / 4 = -q1q2
or
3[q12 + q22 + 2 q1q2 ] = -4q1q2
or
3q12 + 3q22 + 10q1q2 ] = 0
now, by diving both sides by q22, 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