Question

Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 5.14 nC at the origin. (Let r₁₂ = 0.230 m.)

Three point charges lie along the axes in the x y-coordinate plane.

• Positive charge q is at the origin.
• A charge of 6.00 nC is at (r_{1 2}, 0), where r_{1 2} > 0.
• A charge of −3.00 nC is at (0, −0.100 m).

1) what is the magnitude and direction?

magnitude	N
direction	° counterclockwise from the -x (negative x) axis

Answer 1

The force directions are shown below. F, The force to -3.00 nC is, - kq)(-3x10 F, =- (0.100) (9x10°)(5.14x10 ")(-3.00x10°) (0.100 =-13.878x10-N = 13.878 UN The force die to 6.00 nC is, k(q)(6.00x10) (0.230) (9x10(5.14x10-) (6.00x10°) (0.230) = 5.247x100N =5.247 UN The net force is, F= F + F1 = (-13.878*10*) +(5.247x10-62 = 14.8x10-N =14.8 UN The direction is, = tan-1 5.247x10-5) (-13.878x106 = -20.7