From figure , for the equilibrium of a particle along a verival line, we get
`T cos theta=mg` ...(i)
While for equilibrium in the plane of equilateral tringal. We get.
`T sin theta=2F cos 30^(@)`...(ii)
So from Eqs.(i) and (ii) we have
`tan thata=(sqrt(3)F)/(mg)`
Here,
`F=(1)/(4 pi epsilon_(0))(q^(2))/(a^(2))` and `tan theta = (OA)/(OP)=(OA)/(sqrt(L^(2)-OA^(2)))`
also from fig, we get
`OA= 2/3AD=2/3 a sin 60^(@)=(a)/(sqrt(3))`
so, `tan theta =(a//sqrt(3))/(sqrt(L^(2)(a^(3)//3)))=(a)/(sqrt(3)L) (as L gtgta)`...(iii)
On substituting the above value of F and tan `theta` in eq(iii) we get
`(a)/(sqrt(3)L)=(sqrt(3))/(mg) (q^(2))/(4 pi epsilon_(0)a^(2))`
i.e `q=[(4 pi epsilon_(0)a^(3)mg)/(3L)]^(1//2)`.