Question

A 1.0-mm-diameter ball bearing has 2.0 * 109 excess electrons. what is the ball bearing's potential?

Answer 1

Bearing potential of Ball= 5760 V

Explanation:

the electric potential is given by

$V=\frac{K Q}{r}$

Q= electrostatic constant= 9 x 10⁹ Nm²/C

Q= charge= n e

n= number of electrons= 2 x 10⁹

e= charge of electron=-1.6 x 10⁻¹⁹ C

so Q=2 x 10⁹ (-1.6 x 10⁻¹⁹ )

Q= -3. 2 x 10⁻¹⁰ C

r= radius of sphere=1/2 mm= 0.0005 m

$V=\frac{9\times10^{9}\times(-3.2\times10^{-10)}}{0.0005}$

V= 5760 volt