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 the charge of a proton and electron

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

Step2: Calculate the net - charge for part (a)

The total charge due to electrons $Q_e = n_eq_e$, where $n_e = 6.15\times10^{6}$, and the total charge due to protons $Q_p=n_pq_p$, where $n_p = 7.44\times10^{6}$.
The net charge $Q = Q_p+Q_e=n_pq_p + n_eq_e$.
Substitute the values:
\[

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

\]

Step3: Calculate the net - charge for part (b)

Here, $n_e = 212$ and $n_p = 165$.
The net charge $Q=n_pq_p + n_eq_e=(165)\times(1.6\times10^{-19}\ C)+(212)\times(-1.6\times10^{-19}\ C)$.
\[

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

\]

Answer:

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