Two point charges placed at a certain distance $r$ in air exert a force $F$ on each other. Then the distance $r'$ at which these charges will exert the same force in a medium of dielectric constant $k$ is given by

Force between two point charges $q_1$ and $q_2$ placed in vacuum at ' $r$ ' $\mathrm{cm}$ apart is $F$. Force between them when placed in a medium having dielectric $\mathrm{K}=5$ at $\mathrm{r} / 5$ $\mathrm{cm}$ apart will be:

A charge of $4\,\mu C$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be.

Two point charges $ + 9e$ and $ + e$ are at $16\, cm$ away from each other. Where should another charge $q$ be placed between them so that the system remains in equilibrium

In a medium, the force of attraction between two point charges, distance $d$ apart, is $F$. What distance apart should these point charges be kept in the same medium, so that the force between them becomes $16\, F$ ?

$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$