Question

A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.5 µC, m = 25 mg, r1 = 1.3 mm, and r2 = 3.6 mm.

Answer 1

When the particles are held stationary, they have Potential Energy bewteen them. Since they are the same sign, they will repel each other.

When the one charge is released, as it move away, it will lose PE and gain KE so that total energy remains the same, thus...

PE_i = PE_f + KE

PE = kqq/r, so...

kqq/r₁ = kqq/r₂ + .5mv²   (mass must be in kg. 25 mg = 2.5 X 10^-5 kg)

(9 X 10⁹)(3.5 X 10^-6)(3.5 X 10^-6)/(.0013) = (9 X 10⁹)(3.5 X 10^-6)(3.5 X 10^-6)/(.0036) + (.5)(2.5 X 10^-5)(v²)

Solve for v,

v=2081.97 m/s