A rigid insulated wire frame in the form of a right-angled triangle ABC is set in a vertical plane as shown in fig. two beads of equal masses m each and carrying charges `q_(1)` and `q_(2)` are connected by a cord of length l and can slide without friction on the wires.
Cinsidering the case when the beads are stationary, determine
(i) the angel `alpha`
(ii) the tension in the cord, and
(iii) The normal recation on teh beads.
If the cord is now cut, what are the value of the charges for which the beads continue to remain stationary?

Explanation

Tension and electrostatic force are in opposite directions and along the string. Now each bead is in equilibirum under the concurrent forces-normal reaction (N), weight (mg), and the resultant of tension and electrostatic force i.e., `T-F_(e)`, where
`F_(e)=(1)/(4 pi epsilon_(0)) (q_(1)q_(2))/(l^(2))`

Applying Lami's theorem the beads, we get
`(N_(1))/(sin(120-alpha))=(mg)/(cos alpha)=(T-F_(e))/(cos 60^(@))`...(i)
`(N_(2))/(sin(60^(@)_alpha))=(mg)/(sin alpha)=(mg)/(sin alpha)=(T-F_(e))/(cos 30^(@))`...(ii)
Divinding Eq. (i) by Eq.(ii) we have
(i) `tan alpha=(cos 30^(@))/(cos 60^(@))=sqrt(3)` or `alpha =60^(@)`
(ii) `T=F_(e)+mg=((1)/(4 pi epsilon_(0)))(q_(1)q_(2))/(l^(2))+mg` ...(iii)
(iii) `N_(1)=sqrt(3)mg` and `N_(2)=mg`
From Eq. (iii), T=0 when the string is cut
or, `q_(1)q_(2)=-(4 pi epsilon_(0))mg l^(2)`