If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

Two identical spheres each of radius $R$ are kept at center-to-center spacing $4R$ as shown in the figure. They are charged and the electrostatic force of interaction between them is first calculated assuming them point like charges at their centers and the force is also measured experimentally. The calculated and measured forces are denoted by $F_c$ and $F_m$ respectively.
($F_c$ and $F_m$ denote magnitude of force)

The plates $S$ and $T$ of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is: [Area of plates = $A$]

Why Coulomb’s law is associated with Newton’s $3^{rd}$ law ?

The charges on two sphere are $+7\,\mu C$ and $-5\,\mu C$ respectively. They experience a force $F$. If each of them is given and additional charge of $-2\,\mu C$, the new force of attraction will be

A certain charge $Q$ is divided into two parts $q$ and $(Q-q) .$ How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?