Two identical pendulum $A$ and $B$ are suspended from the same point. The bobs are given positive charges, with $A$ having more charge than $B$ . They diverge and reach at equilibrium, with $A$ and $B$ making angles $\theta _1$ and $\theta _2$ with the vertical respectively, Then

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

When the distance between the charged particles is halved, the force between them becomes

A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to

The charges on two sphere are $+7\,\mu C$ and $-5\,\mu C$ respectively. They experience a force $F$. If each of them is given and additional charge of $-2\,\mu C$, the new force of attraction will be

A charged particle having some mass is resting in equilibrium at a height $H$ above the centre of a uniformly charged non-conducting horizontal ring of radius $R$. The force of gravity acts downwards. The equilibrium of the particle will be stable $R$