${F_g}$ and ${F_e}$ represents gravitational and electrostatic force respectively between electrons situated at a distance $10\, cm$. The ratio of ${F_g}/{F_e}$ is of the order of

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r.$ Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become

Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$. What is the equilibrium separation when each thread makes a small angle $\theta$ with the vertical?

Two identical spheres each of radius $R$ are kept at center-to-center spacing $4R$ as shown in the figure. They are charged and the electrostatic force of interaction between them is first calculated assuming them point like charges at their centers and the force is also measured experimentally. The calculated and measured forces are denoted by $F_c$ and $F_m$ respectively.
($F_c$ and $F_m$ denote magnitude of force)

Given below are three schematic graphs of potential energy $V(r)$ versus distance $r$ for three atomic particles : electron $\left(e^{-}\right)$, proton $\left(p^{+}\right)$and neutron $(n)$, in the presence of a nucleus at the origin $O$. The radius of the nucleus is $r_0$. The scale on the $V$-axis may not be the same for all figures. The correct pairing of each graph with the corresponding atomic particle is

Two positively charged spheres of masses $m_1$ and $m_2$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$. Charges on the two spheres are $q_1$ and $q_2$, respectively. At equilibrium, both strings make the same angle $\theta$ with the vertical. Then