Two identical charges repel each other with a force equal to $10 \text{ mg}$ wt when they are $0.6 \text{ m}$ apart in air $(g = 10 \text{ ms}^{-2})$. The value of each charge is:

$A$ paisa coin is made up of $Al-Mg$ alloy and weighs $0.75\,g$. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\,kC$. Suppose that these equal charges were concentrated in two point charges separated by:
$(i)$ $1\,cm$ $(\sim \frac{1}{2} \times \text{diagonal of the one paisa coin})$
$(ii)$ $100\,m$ $(\sim \text{length of a long building})$
$(iii)$ $10^6\,m$ $(\text{radius of the earth})$.
Find the force on each such point charge in each of the three cases. What do you conclude from these results?

$A$ particle of mass $M$ and charge $q$ is placed at the midpoint between two fixed charges each of magnitude $Q$, separated by a distance $2d$. The system is collinear as shown in the figure. If the particle is displaced by a small distance $x$ $(x \ll d)$ along the line joining the two charges and released, it will oscillate about the mean position with a time period $T$. ($\varepsilon_{0}$ is the permittivity of free space)

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is $r$. Now,the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes: