Source: High school physics (Chinese)
Problem
The elementary charge is $e \equiv 1.602\,176\,634 \times 10^{-19}\,\text{C},$.
The number of excess electrons $N$ in a total charge $Q$ is given by:
$$N = -\frac{Q}{e}$$For $Q = -1\mu\text{C}$, the number of excess electrons is:
$$N \approx 6.24 \times 10^{12} \text{ electrons}$$The principle of charge quantization states that any total charge $Q$ is an integer multiple of the elementary charge $e$. The charge of a single electron is $q_e = -e$. If a net charge $Q$ is due to an excess of $N$ electrons, the total charge is the sum of the charges of these electrons.
The relationship between the total charge $Q$, the number of excess electrons $N$, and the elementary charge $e$ is:
$$Q = N q_e = N(-e)$$To find the number of excess electrons $N$, we rearrange this equation:
$$N = -\frac{Q}{e}$$This is the general expression for the number of excess electrons. We can now substitute the given values: $Q = -1\mu\text{C} = -1 \times 10^{-6}\,\text{C}$ and $e = 1.602 \times 10^{-19}\,\text{C}$.
$$N = -\frac{-1 \times 10^{-6}\,\text{C}}{1.602 \times 10^{-19}\,\text{C}}$$ $$N = \frac{1}{1.602} \times 10^{13}$$ $$N \approx 6.24 \times 10^{12}$$Since $N$ must be an integer, this result represents an extremely large number of electrons.