Two charges`+q_(1)` and `-q_(2)` are placed at A and B respectively. A line of force emanates from `q_(1)` at an angle `alpha` with the line AB. At what angle will it terminate at `-q_(2)` ?

Answer

`2 sin^(-1)[("sin"alpha/2)sqrt(q_(1)/q_(2))]`

Explanation

Solid angle `=2pi (1- cos alpha)`
`q_(1)/(2 in_(0))(1- cos alpha)=q_(2)/(2 in_(0))(1- cos theta)`
`q_(1) (sin^(2) alpha)/2=q_(2)(sin^(2) theta)/2`
`"sin" theta/2=sqrt(q_(1)/q_(2)) "sin" alpha/2`
`theta=2 sin^(-1)[sqrt(q_(1)/q_(2))"sin" alpha/2]`