R = 0.30 m and V = 50 m/s.
If V^2/R = k Q q/R^2 , there will be circular motion.
Solve for Q
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.690 g, q = 4.75 µC) is located on the x axis at x = 20.0 cm, moving with a speed of 50.0 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
2 answers
Here R = 0.20 m
mV^2 /R = kQq/R^2
Am I right?
mV^2 /R = kQq/R^2
Am I right?
2 answers