$A$ 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:

$A$ charge $+Q$ is placed at each of the diagonally opposite corners of a square. $A$ charge $-q$ is placed at each of the other diagonally opposite corners as shown. If the net electrical force on $+Q$ is zero,then $\frac{+Q}{-q}$ is equal to

Two identical balls having like charges $Q_1$ and $Q_2$ are placed at a distance $r$ apart and repel each other with a force $F$. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls, $\frac{Q_1}{Q_2}$, is:

Charges $3Q$,$q$,and $Q$ are placed along the $x$-axis at positions $x=0$,$x=l/3$,and $x=l$ respectively. When the net force on charge $Q$ is zero,the value of $q$ is:

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium,assuming that the angle made by each thread is small,the separation $x$ between the balls is proportional to

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is