An amoeba has $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{16}}$ protons and a net charge of 0.300 pC . (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

• Net charge on amoeba is$0.300pC$.

When the number of electrons is less than the number of protons in the body, the body is said to be positively charged.

According to quantization of charge,

$Q=n\left|q_e\right|$

Here, Q is the net charge on the body $\left(Q=0.300pC\right)$, is the excess number of protons responsible for the net charge, and $\left|q_e\right|$is the fundamental unit of charge

$\left(\left|q_e\right|=1.6\times {10}^{-19}C\right)$

Therefore, the excess number of protons responsible for net charge is,

$n=\frac{Q}{\left|q_e\right|}$

Substituting all known values,

$$\begin{gathered} n=\frac{\left(0.300pC\right)}{\left(1.6\times {10}^{-19}C\right)} \\ =\frac{\left(0.300pC\right)\times \left({\displaystyle \frac{{10}^{-12}C}{1pC}}\right)}{\left(1.6\times {10}^{-19}C\right)} \\ =1.875\times {10}^6 \end{gathered}$$

Hence, there are $1.875\times {10}^6$fewer electrons than protons in an amoeba.

The fraction of protons that would have no electrons is,

$f=\frac{n}{n_p}$

Here, n is the number of protons that would have no electrons $\left(n=1.875\times {10}^6\right)$ , and n_p is the total number of protons $\left(n_p=1.00\times {10}^{16}\right)$.

Substituting all known values,

$$\begin{gathered} f=\frac{1.875\times {10}^6}{1.00\times {10}^{16}} \\ =1.875\times {10}^{-10} \end{gathered}$$

Hence, the fraction of protons would have no electrons is $\frac{1}{1.875\times {10}^{10}}$.