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))sin theta`

Answer

c

Explanation

`F_(2)=` Force applied by `q_(2)` on `q_(1)`
`F_(3)=` Force applied by `(-q_(3))` on `-q_(1)`
`x`-component of Net force on `-q_(1)` is
`F_(x)=F_(2)+F_(3) sin theta=k(q_(1)q_(2))/(b^(2))+k.(q_(1)q_(3))/(a^(2))sin theta`
`implies F_(x)=k[(q_(1)q_(2))/(b^(2))+(q_(1)q_(3))/(a^(2))sin theta]`
`implies F_(x)=k.q_(1)[(q_(2))/(b^(2))+(q_(3))/(a^(2))sin theta]`
`implies F_(x) prop((q_(2))/(b^(2))+(q_(3))/(a^(2))sin theta)`