A copper atom has 29 electrons revolving around the nucleus. A copper ball contains $4\times {10}^{23}$ atoms. What fraction of the electron be removed to give the ball a charge of $+9.6\mathrm{\mu C}$?

Here , No of electrons = 29 

Total no of atoms = 4 x 10 ²³

⇒ Total number of electrons ( N) = 29 x 4 x 10²³ = 116 x 10²³

Charge of the ball (q)= 9.6 μ C $\Rightarrow$the number of electrons = 6x10¹³

fraction to be removed is $\frac{6x {10}^{13}}{116x {10}^{23}}\simeq 5.2 x{10}^{-12}$