$A$ charge $Q$ is placed at each of the opposite corners of a square. $A$ charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero,then $Q/q$ equals:

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, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges $Q$ repel each other with a force $F$ when kept apart at a distance $r$. $A$ third spherical conductor having the same radius as that of $B$ but uncharged is brought in contact with $B$,then brought in contact with $C$,and finally removed away from both. The new force of repulsion between $B$ and $C$ is

$A$ charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum,is

Two identical conducting spheres with negligible volume have $2.1 \, nC$ and $-0.1 \, nC$ charges,respectively. They are brought into contact and then separated by a distance of $0.5 \, m$. The electrostatic force acting between the spheres is $.......... \times 10^{-9} \, N$. [Given: $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, N \cdot m^{2}/C^{2}$]

Why is the Coulombian force called a two-body force?