Question

How many electrons would have to be removed from a coin to leave it with a charge of $\mathbf{+}\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{\text{C}}$ ?

Short answer

The number of electrons to be removed is$6.3\times {10}^{11}.$

Step-by-step solution

The given data 

Charge of the coin, $\mathrm{q}=+1.00\times {10}^{-7}\mathrm{C}$

Understanding the concept of quantization of charge 

As we know that a charged particle emits energy in the form of packets, thus, the number of electrons emitted or present can be considered as multiple factors of the electronic charge.

Formula:

The number of electrons present, $\mathrm{n}=\mathrm{q}/\mathrm{e}$ (1)

Calculation of the number of electrons

Using the equation (1) and the given data, we can get the number of electrons removed from the coin as given:

$\begin{array}{c}\mathrm{n}=\frac{1.00\times {10}^{-7}\mathrm{C}}{1.60\times {10}^{-19}\mathrm{C}}\\ =6.3\times {10}^{11}\end{array}$

Hence, the number of electrons removed is $6.3\times {10}^{11}$