Please help ASAP

Question

Charge q1 = 5.01·10–8 C is placed at the origin. Charges q2 = -1.95·10–8 C and q3 = 2.03·10–8 C are placed at points (0.151 m,0 m) and (0 m,0.323 m), respectively, as shown in the figure. Determine the direction of the net electrostatic force on charge q3 (angle with respect to the positive x-axis)

Answer 1

Take k below as 8.987551788E9 Newton-Dot-Square-Meter Per Square Coulomb 

(or N•m²/C²).

Force from q₁ on q₃ is k(5.01E-8 m)(2.03E-8 m) divided by (0.323 m)² which goes to

8.761331744E-5 Newton at 90°.

Force from q₂ on q₃ is k(1.95E-8 m)(2.03E-8 m) divided by (√0.12713 m)² which goes to

2.798491603E-5 Newton at [270 + Arctan (151/323)]°.

A Force of 8.761331744E-5 Newton at 90° (or going "straight up") added to a Force of 2.798491603E-5 Newton at 295.0557249° (or going "down and to the right") gives a final Force & Direction to q₃ of 6.33798287E-5 Newton at 79.22263603° (or going "up and to the right").

• q₃ = 2.03E-8 C at (0, 0.323 m)

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•------------------•q₂ = -1.95E-8 C at (0.151 m, 0)

q₁ = 5.01E-8 C at (0, 0)