$A$ charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. The force of repulsion between them will be maximum when:

Two point charges $q_1$ and $q_2$ are $l$ distance apart. If one of the charges is doubled and the distance between them is halved,the magnitude of force becomes $n$ times,where $n$ is

If the Coulombian force acting between two protons separated by a distance $r$ is $F$, what would be the force acting between two alpha particles separated by a distance $2r$?

Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x$ $(x \ll a)$ along $X$-axis. The force acting on the point charge is ($x$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

The Coulombian repulsive force between two alpha particles kept at a distance of $3 \ cm$ in air is . . . . . . .

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?