Question

A point charge Q₁ = +4.3 μC is fixed in space, while a point charge Q₂ = +2.9 nC, with mass 7.8 μg, is free to move around nearby.

B) If Q₂ is released from rest at a point 0.36 m from Q₁, what will be its speed, in meters per second, when it is 0.84 m fromQ₁?

Answer 1

Q₁ = 4.3 x 10^-6 C

Q₂ = 2.9 x 10^-9 C

m = mass of charge Q₂ = 7.8 x 10^-6 g = 7.8 x 10^-9 kg

r_i = initial distance between the charges where Q2 is released = 0.36 m

r_f = final distance between charges .= 0.84 m

v = speed of charge Q₂ at final distance

Using conservation of energy

initial electric potential energy = final electric potential energy + kinetic energy of charge Q₂

k Q₁ Q₂/r_i = k Q₁ Q₂/r_f + (0.5) m v²

(9 x 10⁹) (4.3 x 10^-6) (2.9 x 10^-9)/(0.36) = (9 x 10⁹) (4.3 x 10^-6) (2.9 x 10^-9)/(0.84) + (0.5) (7.8 x 10^-9) v²

v = 213.72 m/s