$A$ 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?

Write the limitations of Coulomb's law.

Three charges $+5 q, Q$ and $-2 q$ are kept along a straight line in the same order such that $+5 q$ and $-2 q$ charges are at a distance of $\frac{2 r}{3}$ and $\frac{r}{3}$ from the charge $Q$ respectively. If the net force on the charge $-2 q$ is zero,then $Q$ is

Electrostatic force between two identical charges placed in vacuum at a distance of $r$ is $F$. $A$ slab of width $\frac{r}{5}$ and dielectric constant $9$ is inserted between these two charges. The new force between the charges is:

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant $K = \frac{1}{4 \pi \varepsilon_0}$ to the gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $= 1.6 \times 10^{-19} \; C$,the mass of the electron $= 9.11 \times 10^{-31} \; kg$,the mass of the proton $= 1.67 \times 10^{-27} \; kg$):