$A$ 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:

$A$ charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. The force of repulsion between them will be maximum when:

In a hydrogen-like system,the ratio of the Coulomb force to the gravitational force between an electron and a proton is of the order of:

Two balls with equal charges are in a vessel with ice at $-10^{\circ} C$ at a distance of $25 \ cm$ from each other. On forming water at $0^{\circ} C$,the balls are brought nearer to $5 \ cm$ for the interaction between them to be same. If the dielectric constant of water at $0^{\circ} C$ is $80$,the dielectric constant of ice at $-10^{\circ} C$ is:

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

The sum of two point positive charges separated by a distance of $1.5 \ m$ in air is $25 \mu C$. If the electrostatic force between the two charges is $0.6 \ N$,then the difference between the two charges is (in $\mu C$)