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

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

$A$ certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge on $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

Positive point charges are placed at the vertices of a star shape as shown in the figure. The direction of the electrostatic force on a negative point charge at the centre $O$ of the star is

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to: