Question

1. find the net charge of a system consisting of (a) 6.15 × 10^6 electrons and 7.44 × 10^6 protons and (b) 212 electrons and 165 protons.

Explanation:

Step1: Recall charge of electron and proton

The charge of an electron $e=- 1.6\times10^{-19}\ C$ and the charge of a proton $q = 1.6\times10^{-19}\ C$.

Step2: Calculate net - charge for part (a)

The total charge of electrons $Q_e=a\times e$, where $a = 6.15\times10^{6}$, and the total charge of protons $Q_p=b\times q$, where $b = 7.44\times10^{6}$.
The net charge $Q = Q_p+Q_e=(b\times q)+(a\times e)$.
Substitute the values:
\[

$$\begin{align*} Q&=(7.44\times 10^{6}\times1.6\times10^{-19})+(6.15\times 10^{6}\times(-1.6\times10^{-19}))\\ &=(7.44 - 6.15)\times10^{6}\times1.6\times10^{-19}\\ &=1.29\times10^{6}\times1.6\times10^{-19}\\ &=2.064\times10^{-13}\ C \end{align*}$$

\]

Step3: Calculate net - charge for part (b)

Let the number of electrons $m = 212$ and the number of protons $n = 165$.
The net charge $Q'=n\times q+m\times e$.
\[

$$\begin{align*} Q'&=(165\times1.6\times10^{-19})+(212\times(-1.6\times10^{-19}))\\ &=(165 - 212)\times1.6\times10^{-19}\\ &=- 47\times1.6\times10^{-19}\\ &=-7.52\times10^{-18}\ C \end{align*}$$

\]

Answer:

(a) $2.064\times10^{-13}\ C$
(b) $-7.52\times10^{-18}\ C$