Find the net force exerted on Q1 due to Q2 and Q3.

Question

Three point charges Q1,Q2 , and Q3 with magnitudes -4.12μC  , 3.74μC  , and -1.14μC , respectively, are located along the x-axis at  0m  , 0.580m  , and 0.900m , respectively. Find the net force exerted on Q1 due to Q2  and Q3.

Answer 1

Starting with our equations:

F = kQ₁Q₂/r²

k = 9*10⁹

The Principle of Superposition allows us to add the vectors of forces to find the net force on Q1

F₁ = F_1,2 + F_1,3

Solving for each force we get

F_1,2= 9*10⁹ x (-4.12 x 10^-6 C) x (3.74 x 10^-6 C) / (.580m)² = -.412 N (+x direction as opposites attract)

F_1,3= 9*10⁹ x (-4.12 x 10^-6 C) x (-1.14 x 10^-6 C) / (.900m)² = .0522 N (-x direction as likes repel)

F₁ = -.412 N + .0522 N = -.360 N

In other words, there are .360N of force acting on Q1 in the positive X direction

Hope this helps!