Write limitation of Coulomb’s law.

Four point charges, each of $+ q$, are rigidly fixed at the four corners of a square planar soap film of side ' $a$ ' The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k\left[\frac{q^2}{\gamma}\right]^{1 / N}$, where ' $k$ ' is a constant. Then $N$ is

Two equal positive point charges are separated by a distance $2 a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is $................$

As shown in the figure. a configuration of two equal point charges $\left( q _0=+2 \mu C \right)$ is placed on an inclined plane. Mass of each point charge is $20\,g$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $h = x \times 10^{-3}\,m$ The value of $x$ is $..........$.(Take $\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\,Nm ^2 C ^{-2}, g=10\,ms ^{-1}\right)$

Two point charges $ + 3\,\mu C$ and $ + 8\,\mu C$ repel each other with a force of $40\,N$. If a charge of $ - 5\,\mu C$ is added to each of them, then the force between them will become....$N$

A particle of mass $1 \,{mg}$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $'2 \,{m}^{\prime}$ when each is carrying same charge $'q'.$ If the free charged particle is displaced from its equilibrium position through distance $'x'$ $(x\,< \,1\, {m})$. The particle executes $SHM.$ Its angular frequency of oscillation will be $....\,\times 10^{8}\, {rad} / {s}$ if ${q}^{2}=10\, {C}^{2}$