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