Four charges are placed at the four corners of a square of side 15 cm. The charges on the upper left and right corners are +3 µC and -6 µC respectively. The charges on the lower left and right corners are -2.4 µC and -9 µC respectively. The net electric force on -6 µC charge is:

Question

a) 25 N, 780 above the -ve x- axis

b) 25 N, 780 below the -ve x- axis

c) 13 N, 310 below the +ve x- axis

d) 18 N, 750 above the +ve x- axis

Thanks!!!!!!

Answer 1

F = k q1 q2 /r^2

all on 6
F = (k q6) q/r^2

let d = .15 meter

due to 3
Fx = - (k q6) 3/d^2 negative direction because +and -
Fy = 0

due to -2.4 repelling
Fx = +(kq6) 2.4 /[ 2*(d)^2](sqrt2/2)
Fy = same up

due to -1 only up
Fy = (kq6)9/ d^2

so
Fx = -c 3/d^2 + c 2.4 *sqrt 2/4d^2

Fy = +c 9 /d^2 + c 2.4 *sqrt 2/4d^2

Fx = (c/d^2) [(2.4/4) sqrt 2 -3]
Fy = (c/d^2) [(2.4/4) sqrt 2+9 ]

tan angle = Fy/Fx

= [.848 + 9]/[.848 -3]

= [ 9.848 /-2.15]

= - 4.58
overall being pushed up and pulled left so
so angle = 77.7 degree above -x axis
I will leave you to do the magnitude
but it is about 78 degrees above the -x axis :)
so if by 780 you mean 78, the answer is a