Question

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is located on the x axis at x = 19.3 cm, moving with a speed of 49.2 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

Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

$Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}$

Solving for Q:

$Q = \frac{m*V^{2}*r}{K*q}$

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC