When air is replaced by a dielectric medium of constant $k$,the maximum force of attraction between two charges separated by a distance $r$:

If $\varepsilon_0$ and $\varepsilon$ represent the permittivity of free space and absolute permittivity of a medium,then the relative permittivity of the medium is:

Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:

What is the force between two small charged spheres having charges of $2 \times 10^{-7} C$ and $3 \times 10^{-7} C$ placed $30 cm$ apart in air?

$A$ paisa coin is made up of $Al-Mg$ alloy and weighs $0.75\,g$. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\,kC$. Suppose that these equal charges were concentrated in two point charges separated by:
$(i)$ $1\,cm$ $(\sim \frac{1}{2} \times \text{diagonal of the one paisa coin})$
$(ii)$ $100\,m$ $(\sim \text{length of a long building})$
$(iii)$ $10^6\,m$ $(\text{radius of the earth})$.
Find the force on each such point charge in each of the three cases. What do you conclude from these results?

The repulsive force between two particles of same mass and charge,separated by a certain distance,is equal to the weight of one of them. The distance between them is . . . . . . $\times 10^{-1} \ m$.
Mass of particle $= 1.66 \times 10^{-27} \ kg$
Charge of particle $= 1.6 \times 10^{-19} \ C$
$k = 9 \times 10^9 \ MKS, \ g = 10 \ ms^{-2}$