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)` ?

Explanation

We know that number of lines of force emerge is proportional to magnitude of the charge. The field lines emanating from `Q_(1)`, spread out equally in all directions. The number of field lines or flux through cone of half angle `theta` is `(Q_(1))/(4pi) 2pi (1- cos theta)`.
Similarly the number of lines of force terminating on `-Q_(2)` at an angle `phi " is " (Q_(2))/(4pi) 2pi (1- cos phi)`. The total lines of force emanating from `Q_(1)` is equal to the total lines of force terminating on `Q_(2)`
`rArr (Q_(1))/(4pi) 2pi (1- cos theta) = (Q_(2))/(4pi) 2pi (1- cos phi) or (Q_(1))/(2) (1- cos theta) = (Q_(2))/(2) (1- cos phi)`
`Q_(1) sin^(2) theta//2 = Q_(2) sin^(2) phi//2`
`sin phi//2= sqrt((Q_(1))/(Q_(2))) sin theta//2 rArr phi= 2 sin^(-1) {sqrt((Q_(1))/(Q_(2))) sin theta//2}`