Question

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.778 g, q = 5.07 µC is located on the x axis at x = 15.1 cm, moving with a speed of 58.0 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Answer 1

First consider the problem with electrostatic point of view.

The force of attraction between the two charges -

F = k*q1*q2 / r^2 = (9 x 10^9 x 5.07 x 10^-6 x Q) / 0.151^2--------------------------------------(i)

Again when an object rotates in a circular path,

the centripetal force, F = m*v^2 / r = (0.778 x 10^-3 x 58^2) / 0.151---------------------------(ii)

Now, the force of attraction between the charges should be equal to the centripetal force.

Therefore, we should have

(9 x 10^9 x 5.07 x 10^-6 x Q) / 0.151^2 = (0.778 x 10^-3 x 58^2) / 0.151

=> (45.63 x 10^3 x Q) / 0.151 = 17332.40 x 10^-3

=> Q = (17332.40 x 0.151x 10^-3) / (45.63 x 10^3) = 57.36 x 10^-6 C = 57.36 micro C.