$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. These charges are placed at a distance $R$. What will be the values of $Q_1$ and $Q_2$ for the maximum repulsive force between them?

Three charges of equal magnitude are placed at three corners of a square of side $a$. If $F_{12}$ is the force between the charges at corners $1$ and $2$,and $F_{13}$ is the force between the charges at corners $1$ and $3$,what is the ratio $F_{12}/F_{13}$?

$A$ total charge $Q$ is broken into two parts $Q_1$ and $Q_2$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when:

$A$ $10\,\mu C$ charge is divided into two parts and placed at $1\,cm$ distance so that the repulsive force between them is maximum. The charges of the two parts are :

$A$ particle of mass $1 \, mg$ and charge $q$ is placed at the midpoint of two stationary particles,each carrying the same charge $q$,kept at a distance of $2 \, m$. If the free charged particle is displaced from its equilibrium position by a small distance $x$ $(x \ll 1 \, m)$,it executes $SHM$. Its angular frequency of oscillation will be $.... \times 10^{8} \, rad/s$ if $q^{2} = 10 \, C^{2}$.

$A$ point charge $q_1 = 4 q_0$ is placed at the origin. Another point charge $q_2 = -q_0$ is placed at $x = 12 \, cm$. The charge of a proton is $q_0$. $A$ proton is placed on the $x$-axis such that the net electrostatic force on it is zero. In this situation,the position of the proton from the origin is $.......... \, cm$.