The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $= 6.7 \times 10^{-11} \, Nm^2/kg^2$,mass of an electron $= 9.1 \times 10^{-31} \, kg$,charge on an electron $= 1.6 \times 10^{-19} \, C$).

Three equal charges are placed on the three corners of a square as shown below. If the magnitude of force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,then the ratio of $F_{13}$ to $F_{12}$ is

In the given diagram,find the distance of the neutral point from the particle of charge $e$ in $cm$.

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:

Two charges $Q$ are placed at opposite corners of a square. Two charges $q$ are placed at the other two corners. If the net electric force on $Q$ is zero,then $Q/q$ is equal to:

Two point charges $+q_{1}$ and $+q_{2}$ are placed a finite distance '$d$' apart. It is desired to put a third charge $q_{3}$ in between these two charges so that $q_{3}$ is in equilibrium. This is