$A$ charge of $4\,\mu C$ is to be divided into two parts. The distance between the two divided charges is constant. What should be the magnitude of the divided charges so that the electrostatic force between them is maximum?

$A$ proton and an anti-proton come close to each other in vacuum such that the distance between them is $10 \, cm$. Consider the potential energy to be zero at infinity. The velocity at this distance will be ........... $\, m/s$.

How did Coulomb determine the law of electric force between two point charges?

Three charges of each magnitude $100 \mu C$ are placed at the corners $A, B$ and $C$ of an equilateral triangle of side $4 \text{ m}$. If the charges at points $A$ and $C$ are positive and the charge at point $B$ is negative,then the magnitude of the total force acting on the charge at $C$ and the angle made by it with $AC$ are:

$N$ point charges are distributed into two groups and separated by a fixed distance. Then the ratio of maximum to minimum forces between the two groups is ($N$ is even and greater than $2$).

Four identical pendulums are made by attaching a small ball of mass $100 \,g$ on a $20 \,cm$ long thread and suspended from the same point. Now,each ball is given charge $Q$,so that balls move away from each other with each thread making an angle of $45^{\circ}$ from the vertical. The value of $Q$ is close to ..............$\mu C$ $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right.$ in $SI$ units $)$