Question

Two particles are fixed on an x axis. Particle 1 of charge 49.0 μC is located at x = -34.5 cm; particle 2 of charge Q is located at x = 33.6 cm. Particle 3 of charge magnitude 43.9 μC is released from rest on the y axis at y = 34.5 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? (a) Number Enter your answer for part (a) in accordance to the question statement Units Choose the answer for part (a) from the menu in accordance to the question statement (b) Number Enter your answer for part (b) in accordance to the question statement Units Choose the answer for part (b) from the menu in accordance to the question statement.

Answer 1

a)

r₁ = distance of q₁ from q₃ = sqrt(0.345² + 0.345²)

r₂ = distance of q₂ from q₃ = sqrt(0.345² + 0.336²)

For acceleration to be in positive x-direction, net force along y-direction must be zero. hence

F₁ Sin = F₂ Sin

(k q₁ q₃/r₁² ) (0.345/r₁) = (k q₂ q₃/r₂² ) (0.345/r₂)

(q₁ /r₁² ) (0.345/r₁) = (q₂ /r₂² ) (0.345/r₂)

((49 x 10^-6)/(0.345² + 0.345²)) (0.345/sqrt(0.345² + 0.345²)) = (Q/(0.345² + 0.336²)) (0.345/sqrt(0.345² + 0.336²))

Q = 47.11 x 10^-6 C

The sign of the charge would be negative

b)

r₁ = distance of q₁ from q₃ = sqrt(0.345² + 0.345²)

r₂ = distance of q₂ from q₃ = sqrt(0.345² + 0.336²)

For acceleration to be in positive y-direction, net force along x-direction must be zero. hence

F₁ Cos = F₂ Cos

(k q₁ q₃/r₁² ) (0.345/r₁) = (k q₂ q₃/r₂² ) (0.336/r₂)

(q₁ /r₁² ) (0.345/r₁) = (q₂ /r₂² ) (0.336/r₂)

((49 x 10^-6)/(0.345² + 0.345²)) (0.345/sqrt(0.345² + 0.345²)) = (Q/(0.345² + 0.336²)) (0.336/sqrt(0.345² + 0.336²))

Q = 48.4 x 10^-6 C

The sign of the charge would be positive