Question

A 0.700 mm -diameter ball bearing has 2.10×109 excess electrons. What is the ball bearing's potential?

Answer 1

e = magnitude of charge on electron = 1.6 x 10^-19 C

n = number of excess electrons = 2.10 x 10⁹ electrons

Q = magnitude of charge

Magnitude of charge is given as

Q = ne

Q = (2.10 x 10⁹) (1.6 x 10^-19)

Q = 3.36 x 10^-10 C

d = diameter of the ball = 0.700 mm = 0.7 x 10^-3 m

r = radius of the ball = d/2 = (0.7 x 10^-3 )/2 = 3.5 x 10^-4 m

Ball bearing's potential is given as

V = kQ/r

V = (9 x 10⁹) (3.36 x 10^-10)/(3.5 x 10^-4 )

V = 8640 Volts