Please help me with this question's solution

Dear Student,

since the spheres are in touch so the charge divides equally between them, i.e 2 X 10^-6 C

Now, when they seperate out, since the angle between the string's is 60, the angle between the perpendicular bisector of the seperation distance and either string is 30 degrees,


$$\begin{gathered} from geometry of this triangle \\ half the seperation dis\tan ce=l\sin 30 \\ thus total seperation=2l\sin 30 \Rightarrow l \\ where l=string length= 10 cm \\ thus coloumb force acting on either= \frac{k{Q}^2}{l^2}\Rightarrow 3.6 N \\ weight of one sphere is Mg, where mass of sphere is M\left(say\right) \\ in the suspended state, at equilibrium, at either sphere, we have coloumb force acting \\ horizontally outward, and weight acting downward, if we resolve them into components \\ then, \\ by geometry \\ Mg\sin 30=F_C\cos 30 \\ and T=Mg\cos 30+F_C\sin 30 \\ Thus, M=\frac{F_{C}cot30}{g}\Rightarrow \frac{3.6\times 1.73}{10}\Rightarrow \mathbf{0}\mathbf{.}\mathbf{6235}\mathbf{ }\mathit{K}\mathit{g} \end{gathered}$$

Regards.