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

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 ${q}^{2}=10\, {C}^{2}$

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....$N$

In the given figure two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$. Assume that $\theta$ is so small that $\tan \theta \approx \sin \theta $, then for equilibrium $x$ is equal to

If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

Two identical charged spheres suspended from a common point by two massless strings of lengths $l,$ are initially at a distance $d\;(d < < l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v.$ Then $v$ varies as a function of the distance $x$ between the spheres, as