Given,    
 $$\begin{gathered} C\mathrm{harge}, q_1 = 2 \times {10}^{-7}\mathrm{C}, \\ C\mathrm{harge}, {\mathrm{q}}_2 = 3 \times {10}^{-7}\mathrm{C} \\ \mathrm{r} = 30 \mathrm{cm} = 0.3 \mathrm{m} \end{gathered}$$

where, r is the distance between two charges.

Using the formula,

$\therefore$                $\boxed{\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}}$

                          $$\begin{gathered} = \frac{9\times {10}^9\times 2\times {10}^{-7}\times 3\times {10}^{-7}}{(0.3{)}^2} \\ = 6 \times {10}^{-3} \mathrm{N} \left(\mathrm{Repulsive}\right) \end{gathered}$$

Repulsive in nature since both charges are positive.

(a) Given,

   $\mathrm{Force} -\mathrm{F} = 0.2 \mathrm{N}$   
                 $$\begin{gathered} \mathrm{charge} -{\mathrm{q}}_1 = 0.4 \mathrm{\mu C} = 0.4 \times {10}^{-6}\mathrm{C} \\ \mathrm{charge}-{\mathrm{q}}_2 = 0.8 \mathrm{\mu C} = 0.8 \times {10}^{-6}\mathrm{C} \end{gathered}$$   

Now using the formula,                                                                   $\mathrm{F} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Hence,                     $\boxed{{\mathrm{r}}^2 = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{\mathrm{F}}}$
                       $$\begin{gathered} {\mathrm{r}}^2 = \frac{9\times {10}^9\times 0.4\times {10}^{-6}\times 0.8\times {10}^{-6}}{0.2} \\ {\mathrm{r}}^2 = 36 \times 4 \times {10}^{-4} = 144 \times {10}^{-4} \\ \mathrm{r} = 12 \times {10}^{-2}\mathrm{m} = 12 \mathrm{cm}. \end{gathered}$$

where, r is the distance between two spheres.

(b) Force on the second sphere due to the first is same, i.e., 0.2 N because the charges in action are same and force is attractive as charges are unlike in nature.

                       $$\begin{gathered} \mathrm{Electrostatic} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F} = \mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ \mathrm{Gravitational} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F} =\mathrm{G}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2} \end{gathered}$$
                      $$\begin{gathered} \mathrm{where} \mathrm{G} = 6.67 \times {10}^{-11}\mathrm{N} {\mathrm{m}}^2 {\mathrm{kg}}^{-2} \\ {\mathrm{m}}_{\mathrm{e}} = 9.1 \times {10}^{-31} \mathrm{kg} \\ {\mathrm{m}}_{ \mathrm{p}} = 1.67 \times {10}^{-27} \mathrm{kg} \\ \mathrm{e} = 1.6 \times {10}^{-19} \mathrm{C} \end{gathered}$$
and   $\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0} = 9\times {10}^9 \frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}$
Now,       $$\begin{gathered} \frac{{\mathrm{ke}}^2}{{\mathrm{Gm}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}} = \frac{9\times {10}^9{\displaystyle \frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}}\times 1.6\times {10}^{-19}\mathrm{C}\times 1.6\times {10}^{-19}\mathrm{C}}{6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2}\times 9.1\times {10}^{-31}\mathrm{kg}\times 1.67\times {10}^{-27}\mathrm{kg}} \\ = 2 \times 27 \times {10}^{39} \mathrm{which} \mathrm{is} \mathrm{dimensionless}. \end{gathered}$$

It also establishes that the electrostatic force is about 10³⁹ times stronger than the gravitational force.

a) Quantisation of Electric Charges mean the total electric charge(q) of a body is always an integral multiple of a basic quantum charge(e).

i.e.,              $\mathrm{q}=\pm \mathrm{ne}$

Here +e is taken as charge on a proton while –e is taken as charge on an electron. The charge on a proton and an electron are numerically equal i.e., 1.6 x 10^–19 C but opposite in sign.