Three charges `-q_(1), +q_(2)` and `-q_(3)` are placed as shown in the figure. The `x`-component of the force on `-q_(1)` is proportional to

Options

A. `(q_(2))/(b^(2)) -(q_(3))/(a^(2)) sin theta`
B. `(q_(2))/(b^(2)) -(q_(3))/(a^(2))cos theta`
C. `(q_(2))/(b^(2))+(q_(3))/(a^(2))sin theta`
D. `(q_(2))/(b^(2)) +(q_(3))/(a^(2))cos theta`

Answer

A

Explanation

`F_(x) = (kq_(1)q_(2))/(b^(2)) - (kq_(1)q_(3))/(a^(2))sin theta`
`= kq_(1) [(q_(2))/(b^(2))-(q_(3))/(a^(2))sin theta] prop [(q_(2))/(b^(2))-(q_(3))/(a^(2))sin theta]`