The electric field in a certain region is acting radially outward and is given by $E=Ar$ . A charge contained in a sphere of radius $a$ centred at the origin of the field, will be given by:

Two identical charged spheres suspended from a common point by two massless strings of lengths $l$ are initially at a distance $\mathrm{d}(\mathrm{d}<<l)$ 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 $\mathrm{v}$ Then, $\mathrm{v}$ varies as a function of the distance $\mathrm{x}$ between the sphere, as 

Suppose the charge of a proton and an electron differ slightly. One of them is $-e,$ the other is $(e+\mathrm{\Delta }e).$ If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance $d$ (much greater than atomic size) apart is zero, then $\mathrm{\Delta }e$ is of the order of (Given mass of hydrogen, $m_h=1.67\times {10}^{-27} \mathrm{kg})$

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron, in comparison to the time of fall of the proton is