$A$ charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. The force of repulsion between them will be maximum when:

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\overrightarrow{F}$ between the two is (where $K = \frac{1}{4\pi\varepsilon_0}$):

Two spheres having the same radius and mass are suspended by two strings of equal length from the same point in such a way that their surfaces touch each other. On depositing a charge of $2 \times 10^{-6} \ C$ on them,they repel each other such that in equilibrium,the angle between their strings becomes $60^{\circ}$. If the distance from the point of suspension to the center of the sphere is $10 \ cm$,find the mass of each sphere. $(k = 9 \times 10^9 \ SI, g = 10 \ ms^{-2})$. (in $kg$)

If two charges $q_1$ and $q_2$ are separated by a distance '$d$' and placed in a medium of dielectric constant $K$,what will be the equivalent distance between the charges in air for the same electrostatic force?

Two insulated charged metallic spheres $P$ and $Q$ of negligible radii have their centres separated by a distance of $60 \,cm$. The mutual force of electrostatic repulsion, if the charge on each is $6 \times 10^{-7} \,C$, is

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