Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of 0.108 N when . . .

Question : Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of 0.108 N when separated by 50.0 cm, center-to-center. A thin conducting wire then connects the spheres, When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the initial charges on the spheres?


Doubt by Diya

Solution : 
Let the initial charge on the two identical spheres be q1(+ve) and q2(-ve)
F = 0.108 N
r = 50 cm = 0.5 m
According to Coulombs law
F = kq1q2/r2
Fr2/k = q1q2 
q1q2 = [0.108×(0.5)2]/[9×109]
q1q2 = [0.108×0.25]/[9×109]
q1q2 = [0.027×10-9]/9
q1q2 = 0.003×10-9
q1q2 = 3×10-12 — (1)

New charges on the two spheres when they are connected by a thin conducting wires will be

q1'=q2'=(q1-q2)/2
F'=0.0360 N
r'=r=50cm=0.5 m

F'=kq1'q2'/r2
F'r2/k=q1'q2'
[0.0360×(0.5)2]/[9×109]=[(q1-q2)/2][(q1-q2)/2]
[0.0360×0.25]/[9×109]=(q1-q2)2/4
(4×0.009×10-9)/(9)=(q1-q2)2
4×10-12 = (q1-q2)2
q1-q2=2×10-6 — (2)

Using
(q1-q2)2=(q1+q2)2 -4q1q2
(q1+q2)2=(2×10-6)2+4( 3×10-12)
(q1+q2)2 =4×10-12+12×10-12
(q1+q2)2=16×10-12
q1+q2=4×10-6 — (3)

On solving equation (2) and (3)

q1-q2=2×10-6
q1+q2=4×10-6
------------------------
2q1=6×10-6
q1=3×10-6
q1=3µC

putting this in equation (3)
3×10-6 + q2=4×10-6
q2=(4-3)×10-6
q2 =1×10-6
q2=1µC.

∴ q1=3µC (Positive) & q2 = 1µC (Negative)