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)` ?

Options

A. `beta=sin^(-1)(sqrt((q_(1))/(q_(2)))sin""(alpha)/(2))`
B. `beta=2sin^(-1)(sqrt((q_(1))/(q_(2)))sin""(alpha)/(2))`
C. `beta=2sin^(-1)(sqrt((q_(2))/(q_(1)))sin""(alpha)/(2))`
D. `beta=2sin^(-1)(sqrt((q_(1))/(q_(2)))sin""(alpha)/(2))`

Answer

A

Explanation

`q_(1)/(2epsilon_(0))(1-cosalpha)=(q_(2))/(2epsilon_(0))(1-cosbeta)`
`q_(1)2sin^(2)""(alpha)/(2)=q_(2)(2sin^(2)""(beta)/(2))`
`rArr sin^(2)""(beta)/(2)=(q_(1))/(q_(2))sin^(2)""(alpha)/(2)rArr beta=2sin^(2)(sqrt((q_(1))/(q_(2)))sin""(alpha)/(2))`