What is the force between two small charged spheres having charges of 2 x 10^–7 C and 3 x 10^–7C placed 30 cm apart in air?

Given,    
 

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.

                       $$\begin{gathered} \mathrm{Electrostatic} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F} = \mathrm{K}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ \mathrm{Gravitational} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F} =\mathrm{G}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2} \end{gathered}$$
                      $$\begin{gathered} \mathrm{where} \mathrm{G} = 6.67 \times {10}^{-11}\mathrm{N} {\mathrm{m}}^2 {\mathrm{kg}}^{-2} \\ {\mathrm{m}}_{\mathrm{e}} = 9.1 \times {10}^{-31} \mathrm{kg} \\ {\mathrm{m}}_{ \mathrm{p}} = 1.67 \times {10}^{-27} \mathrm{kg} \\ \mathrm{e} = 1.6 \times {10}^{-19} \mathrm{C} \end{gathered}$$
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) 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.

Law of conservation of charge states that total charge on an isolated system of objects always remain conserved.
When a glass rod is rubbed with silk cloth, glass rod becomes positively charged while silk cloth becomes negatively charged and, the amount of positive charge on the glass rod is apparently found to be exactly the same as the negative charge on silk cloth. Since, the measure of charge is same on both, equal amount of charge with opposite nature will cancel out each other. Hence, the total sum of charge on two bodies is zero. Thus, the system of glass rod and silk cloth, which was neutral before rubbing, still possesses no net charge after rubbing. And law of conservation of charge is justified. 

As a consequence of conservation of charge, when two charged conductors of same size and same material carrying charges Q₁ and Q₂ respectively are brought in contact and separated, the charge on each conductor will be $\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_2}{2}.$
This condition, however, does not hold true if the conductors are of different sizes or of different material.
In that case the charges on the conductors will be Q₁^' and Q₂’ respectively, where Q₁ + Q₂ = Q₁‘ + Q₂’.