Assume that you have $10^{23}$ carbon atoms. All the nuclei are placed at the North Pole of the Earth and all the electrons are placed at the South Pole of the Earth. (Radius of Earth = $6400 \ km$). The force between the charges is approximately:

Two point charges $q_1$ and $q_2$ are placed at a distance of $50 \ cm$ from each other in air and interact with a certain force. Now,these charges are placed in an oil with a relative permittivity of $5$. If the force between them remains the same,the distance between them in the oil is ........ $cm$.

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. These charges are placed at a distance $R$. What will be the values of $Q_1$ and $Q_2$ for the maximum repulsive force between them?

$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?