Q1= +6.60 x 10-6 C, Q2= +3.10 x 10-6 C, and Q3= +5.30 x 10-6 C. What is the magnitude and direction of the net force on charge q2?

Question

Q1 is connected to Q2 (left to right) with a distance of 0.350 m. Q2 is connected to Q3 (up-down) with a distance of 0.155 m.

Answer 1

The repulsive force acting on Q2 by Q1 from left to right:

Fx =  (9.0 x 10⁹ N•m²/C²) • (+6.60 x 10^-6 C) • (+3.10 x 10^-6C) / (0.35 m)²

^{= 1.65 N}

The repulsive force acting on Q2 by Q3 from left to right:

Fy =  (9.0 x 10⁹ N•m²/C²) • (+3.10 x 10^-6 C) • (+5.30 x 10^-6C) / (0.155 m)²

^{= 6.15 N}

Magnitude of the net force

F = √(Fx²+Fy²) = √( 1.65 N ²+ 6.15 N ²) = 6.37 N

Direction of the net force

θ = tan^-1 (Fy / Fx) = tan^-1 (6.15 N / 1.65 N) = 75° (above the horizontal line to the right)