Answer
$V_{ball}= 5.76*10^{3}$v
Work Step by Step
Number of excess electrons $= 2.0* 10^{9}$
Thus charge $q$ on the ball $= 2.0* 10^{9}* 1.6*10^{-19}$c
$q = 3.2*10^{-10}$c
Radius $r$ of the ball $=\frac{diameter}{2} = 0.5mm = 5*10^{-4}m$
Therefore Potential of the ball $= \frac{k*q}{r}$
$V_{ball}= \frac{9*10^{9} * 3.2*10^{-10}}{5*10^{-4}}$v
$V_{ball}= 5.76*10^{3}$v
