Correct Answer - C
The four possibel force diagrame are shown in fig. Only the last pocture can result in an electric field the negative x-direction
`q_(1)=-2.00(mu)C, q_(2)= +4.00(mu)C, and q_(2)gt0`
`E_(y)=0=(1)/(4 pi epsilon_(0)) (q_1)/((0.0400m)^(2)) sin theta -(1)/(4 pi epsilon_(0))(q_2)/((0.0300m)^(2)) cos theta`
`q_(2)=9/16 (sin theta)/(cos theta)=9/16 q_(1) (3//5)/(4//5)=(27)/(64) q_(1)=0.843mu C`
`F_(3)=q_(3)E_(x)=q_(3)(1)/(4 pi epsilon_(0))((q_(1))/(0.0016)(4)/(5)+(q_(2))/(0.0009)3/5)=56.25N`.