The electrostatic force of repulsion between two positively charged ions carrying equal charge is 3.7 x 10^–9 N, when they are separated by a distance of 5 . How many electrons are missing from each ion?

Here given,
           $\mathrm{Electrostatic} \mathrm{force} \mathrm{of} \mathrm{repulsion} -\mathrm{F} = 3.7 \times {10}^{-9}\mathrm{N}$

$$\begin{gathered} \mathrm{Let} \mathrm{us} \mathrm{say} \mathrm{charge} \mathrm{is} {\mathrm{q}}_1 = {\mathrm{q}}_2 = \mathrm{q} \\ \mathrm{distance} \mathrm{between} \mathrm{two} \mathrm{charges}-r = 5 \mathrm{\AA } = 5 \times {10}^{-10} \mathrm{m}, \\ \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{electrons} \mathrm{missing} - \mathrm{n} = ? \end{gathered}$$

Using Coulomb's law,
                   
                  $\mathrm{F} = \frac{1}{4\mathrm{\pi } {\in }_0}\frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{{\mathrm{r}}^2}$
         
      $$\begin{gathered} \Rightarrow 3.7 \times {10}^{-9} = 9 \times {10}^9 \frac{\mathrm{qq}}{(5 \times {10}^{-10}{)}^2} \\ \Rightarrow {\mathrm{q}}^2 = \frac{3.7 \times {10}^{-9} \times 25 \times {10}^{-20}}{9 \times {10}^9} \\ = 10.28 \times {10}^{-38} \\ \therefore \mathrm{q} = 3.2 \times {10}^{-19} \mathrm{coulomb} \end{gathered}$$

As           $\mathrm{q} = \mathrm{ne}$
$\therefore$            $\mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}=\frac{3.2 \times {10}^{-19}}{1.6 \times {10}^{-19}} = 2$