Two equally charged particles, held 3.2 x 10^–3 m apart, are released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s² and that of the second to be 9.0 m/s². If the mass of the first particle is 6.3 x 10^–7 kg, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

Given, 

Distance between the charges, r = $3.2\times {10}^{-3} m$
 $$\begin{gathered} \mathrm{Iinitial} \mathrm{acceleration} \mathrm{of} \mathrm{first} \mathrm{particle}, {\mathrm{a}}_1 = 7.0 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Initial} \mathrm{acceleration} \mathrm{of} \mathrm{second} \mathrm{particle}, {\mathrm{a}}_2 = 9.0 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Mass} \mathrm{of} \mathrm{first} \mathrm{particle}, {\mathrm{m}}_1 = 6.3 \times {10}^{-7} \mathrm{kg}, \\ \mathrm{Mas} \mathrm{of} \mathrm{second} \mathrm{particle}, {\mathrm{m}}_2 = ? \end{gathered}$$

a) Since,   ${\mathrm{F}}_1 = {\mathrm{F}}_2$

$\therefore$        ${\mathrm{m}}_1{\mathrm{a}}_1 = {\mathrm{m}}_2{\mathrm{a}}_2$

   $$\begin{gathered} \mathrm{mass} \mathrm{of} \mathrm{second} \mathrm{particle}-{\mathrm{m}}_2 = \frac{{\mathrm{m}}_1{\mathrm{a}}_1}{{\mathrm{a}}_2} \\ = \frac{6.3\times {10}^{-7}\times 7.0}{9.0} \\ = 4.9 \times {10}^{-7} \mathrm{kg} \end{gathered}$$ 

b)As,
                $$\begin{gathered} {\mathrm{F}}_1 = {\mathrm{F}}_2 \\ = \frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} = {\mathrm{m}}_1{\mathrm{a}}_1 \\ = 6.3 \times {10}^{-7} \times 7.0 \\ = 44.1 \times {10}^{-7} \end{gathered}$$ 

$\therefore$             $\frac{9\times {10}^9 {\mathrm{q}}^2}{{\left(3.2 \times {10}^{-3}\right)}^2} = 44.1 \times {10}^{-7}$ 

$\therefore$                $\mathrm{q} = 7.1 \times {10}^{-11}\mathrm{C}.$