Question

Two particles are fixed on an x axis. Particle 1 of charge 51.4 μC is located at x = -22.8 cm; particle 2 of charge Q is located at x = 10.8 cm. Particle 3 of charge magnitude 20.8 μC is released from rest on the y axis at y = 22.8 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

Answer 1

In triangle ABC

₁ = tan^-1(AC/BC) = tan^-1(22.8/22.8) = 45

using pythagorean theorem

AB = sqrt(BC² + AC²) = sqrt((22.8)² + (22.8)²) = 32.3 cm = 0.323 m

In triangle ADC

₂ = tan^-1(AC/CD) = tan^-1(22.8/10.8) = 64.6

AD = sqrt(AC² + CD²) = sqrt((22.8)² + (10.8)²) = 25.23 cm = 0.252 m

F₁ = force by charge Q₁ on charge Q₃ = k Q₁ Q₃/AB² = (9 x 10⁹) (51.4 x 10^-6) (20.8 x 10^-6)/(0.323)²

F₂ = force by charge Q₂ on charge Q₃ = k Q₁ Q₃/AD² = (9 x 10⁹) (51.4 x 10^-6) Q₃/(0.252)²

for the acceleration to be along X-direction , the net force must be along X-direction , net force along Y-direction must be zero

hence F_1y = F_2y

F₁ Sin₁ = F₂ Sin₂

(9 x 10⁹) (51.4 x 10^-6) (20.8 x 10^-6) Sin45/(0.323)² = (9 x 10⁹) (51.4 x 10^-6) Q₃ Sin64.6/(0.252)²

Q₃ = 9.91 x 10^-6 C

b)

for the acceleration to be along Y-direction , the net force must be along Y-direction , net force along X-direction must be zero

hence F_1x = F_2x

F₁ Cos₁ = F₂ Cos₂

(9 x 10⁹) (51.4 x 10^-6) (20.8 x 10^-6) Cos45/(0.323)² = (9 x 10⁹) (51.4 x 10^-6) Q₃ Cos64.6/(0.252)²

Q₃ = 20.8 x 10^-6 C