The unit of permittivity of free space ${\varepsilon _0}$ is

Four electric charges $2 \mu C, Q, 4 \mu C$ and $12 \mu C$ are placed on the $x$-axis at distances $x=0, 1 \ cm, 2 \ cm$ and $4 \ cm$ respectively. If the net force acting on the charge at the origin is zero,then $Q=$ (in $\mu C$)

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

$A$ point charge $q$ is placed at a distance $d$ from one end of a thin non-conducting rod of length $L$ having a charge $Q$ uniformly distributed along its length,as shown in the figure. Find the magnitude of the electric force between them.

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.

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: