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

It is the property of lines of force that their number within a tube remains unchanged, and the number of lines of force is equal to the charge. The lines of force emanating form `q_(1)` unit solid angle are `q_(1)//4 pi`, and the number of lines through the cone of half-angel `alpha` is
`(q_(1))/(4 pi) 2pi (1-cos alpha)`
because the solid angle of a cone is `2pi(1-cos alpha)`. Similarly, the number of lines of force terminating on `-q_(2)` at `beta` is
`(q_(2))/(4 pi) 2 pi(1-cos beta)`
By the property of lines of lines of force
`(q_(1))/(4 pi) 2pi (1-cos alpha)= (q_2)/(4 pi) 2 pi(1-cos beta)`
or,`(q_(1))/(2)2" sin"^(2)(alpha)/(2)=(q^(2))/(2)2" sin"^(2)(beta)/(2)`
or `"sin"(beta)/(2)"sin"(alpha)/(2) sqrt((q_(1))/(q_(2)))`
or `beta=2 sin^(-1)["sin"(alpha)/(2)sqrt((q_(1))/(q_(2)))]`.