Particle A of charge 3.00 x 10-4 C is at the origin, particle B of charge -6.00 x 10-4 C is at (4.00 m, 0)

Question

and particle C of charge 1.00 x 10-4 C is at (0, 3.00 m). Find the magnitude and direction of the
resultant electric force acting on C.

Answer 1

X force on C = + k ( 6 )(1)(10^-8) (4/5)
note 4/5 is cos angle and unlike charges attract)
Y force on C = +k(3)(3)(10^-8) - k(6)(1)(10^-8)(3/5)
magnitude = sqrt (X^2+Y^2)
cos theta = X / Y