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?

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

Assertion : The Coulomb force is the dominating force in the universe.

Reason : The Coulomb force is weaker than the gravitational force.

Three equal charges $+q$ are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude $F(r)=k r$ directed towards the origin, where $k$ is a constant. What is the distance of the three charges from the origin?

Two charges $ + 4e$ and $ + e$ are at a distance $x$ apart. At what distance, a charge $q$ must be placed from charge $ + e$ so that it is in equilibrium

Why Coulombian force is called two body force ?