91Ó°ÊÓ

Quantization of charge means that only those values of charge can exist freely in nature, which is integral multiple of the elementary charge. Mathematically,

$\mathit{Q}\mathbf{=}\mathit{n}\mathit{e}$

Here, n is an integer, and role="math" localid="1653632484896" $e$is the elementary charge.

According to the quantization of charge,

$Q=ne$

Here, Q is the total charge $\left(\mathrm{Q}\mathrm{=}\mathrm{2}\mathrm{.}\mathrm{00}\mathrm{}\mathrm{n}\mathrm{C}\right)$,n is the number of electrons needed to develop charge, ande is the charge in one electron $\left(\mathrm{e}\mathrm{=}\mathrm{-}\mathrm{1}\mathrm{.}\mathrm{6}\mathrm{×}{\mathrm{10}}^{\mathrm{-}\mathrm{19}}\mathrm{C}\right)$.

The number of electrons (n) required to produce the charge (Q) is,

$n=\frac{Q}{e}$

Substituting all known values,

$$\begin{gathered} n=\frac{-2.00nC}{-1.6×{10}^{-19}C} \\ =\frac{-2.00nC×\left({\displaystyle \frac{{10}^{-9}C}{1nC}}\right)}{-1.6×{10}^{-19}C} \\ =1.25×{10}^{10} \end{gathered}$$

Hence,$1.25×{10}^{10}$ electrons are needed to produce a charge of $-2.00nC$.

When one electron is removed from a neutral object it gains a charge equal to the charge of proton. The total charged gained on the object by the removal of the electron is,

Q = np

Here, is the net charge gained on the object by the removal of electron $\left(Q=0.500\mathrm{μ}C\right)$, is the number of electrons removed from the object, and p is the charge gained on the object due to removal of electron $\left(p=1.6×{10}^{-19}C\right)$.

The expression for the number of electrons removed from the object is,

$n=\frac{Q}{p}$

Substituting all known values,

$$\begin{gathered} n=\frac{0.500\mathrm{μ}C}{1.6×{10}^{-19}C} \\ =\frac{0.500\mathrm{μ}C×\left({\displaystyle \frac{{10}^{-6}C}{1\mathrm{μ}C}}\right)}{1.6×{10}^{-19}C} \\ =3.125×{10}^{12} \end{gathered}$$

Hence, an object will gain charge of $0.500\mathrm{μ}C$when $3.125×{10}^{12}$electrons are removed.