Coulomb's Law question

Question

A point charge Q₁ = +6.0 nC is at the point (0.30 m, 0.00 m); a charge Q₂ = -1.0 nC is at (0.00 m, 0.10 m), and a charge Q₃ = +5.0 nC is at (0.00 m, 0.00 m). What are the magnitude and direction of the net force on the +5.0-nC charge due to the other two charges? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Answer 1

Please draw a picture of the locations of the charges.

Let

r₁ = 0.30 m be the distance between Q₁ and Q₃,

r₂ = 0.10 m be the distance between Q₂ and Q₃,

F₁ be the force experienced by Q₃ due to Q₁,

F₂ be the force experienced by Q₃ due to Q₂.

The net force is the vector sum of F₁ and F₂.

Their magnitudes are calculated by the Coulomb's Law

F₁ = kQ₁Q₃/r₁² = 9×10⁹×6×10^-9×5×10^-9/0.3² = 3×10^-6

F₂ = kQ₂Q₃/r₂² = 9×10⁹×1×10^-9×5×10^-9/0.1² = 4.5×10^-6

The force F₁ repulses Q₃ towards the right, and

The force F₂ attracts Q₃ upwards.

The net force will be in the second quadrant at an angle α against the negative x-axis such that

tan(α) = F₂/F₁ = 1.5

α = 56.3°

The magnitude of the net force is obtained by the Pythagorean Theorem

F = √(F₁² + F₂²) = √((3×10^-6)² + (4.5×10^-6)²) = 5.4×10^-6 N