Question

Three paricles, each of mass m and carrying a charge q each, are suspended from a common pointby insulating massless strings each of length L. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side a, calculate the charge q on each particle. Assume `Lgtgta`.

Answer 1

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)`.