$A$ charge $Q$ is placed at the midpoint of the line joining two identical point charges $q$. For what value of $q$ will the system be in equilibrium?

Two positive ions,each carrying a charge $q,$ are separated by a distance $d.$ If $F$ is the force of repulsion between the ions,the number of electrons missing from each ion will be ($e$ being the charge on an electron).

Consider the charges $q, q$,and $-q$ placed at the vertices of an equilateral triangle of side $l$,as shown in the figure. What is the force on each charge?

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes:

Two positive charges of $20 \, C$ and $Q \, C$ are situated at a distance of $60 \, cm$. The neutral point between them is at a distance of $20 \, cm$ from the $20 \, C$ charge. The value of charge $Q$ is: (in $, C$)

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at positions $0, d/3$ and $2d/3$ respectively. In order to make the net force on $q$ be zero, the value of $Q$ should be