Question

Three charges are placed as shown in fig. the magnitude of `q_(1)` is `2.00mu C`, but its sign and the value of the charge `q_(3)` is `+4.00muC`, and the net force on `q_(3)` is entirely in the negative x-direction

As per the condition given in the problem, the sign of `q_(1)` and `q_(2)` will be
A. Charge `q_(1)` is negative.
B. Charge `q_(2)` is positive
C. The magnetidu of charge `q_(2)` is `(27)/(32)muC`
D. The magnitude of net force on charge `q_(3)` is `(45)/(22)mN`.

Answer 1

Correct Answer - A::B::C

For F to be along negative x-axis, `q_(1)` has to be negative while `q_(2)` has to be positive.
also `F_(1) cos 53 = F_(2) cos 37^(2)`
where `F_(1)=(K.q_(1)q_(3))/((4 "cm")^(2))` and `F_(2)=(K.q_(2)q_(3))/((3 "cm")^(2))`
on putting values `q_(2) = (27)/(32) muC`