Bearing potential of Ball= 5760 V
Explanation:
the electric potential is given by
[tex] V=\frac{K Q}{r} [/tex]
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
[tex] V=\frac{9\times10^{9}\times(-3.2\times10^{-10)}}{0.0005} [/tex]
V= 5760 volt
