In a region of space, suppose there exists a uniform electric field $\vec{E}=10 i\left(\frac{ v }{ m }\right)$. If a positive charge moves with a velocity $\vec{v}=-2 \hat{j}$, its potential energy

Work done in moving a positive charge on an equipotential surface is

In a region, electric field varies as $E = 2x^2 -4$ where $x$ is the distance in metre from origin along $x-$ axis. A positive charge of $1\,\mu C$ is released with minimum velocity from infinity for crossing the origin, then

There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R(t)$ is

Charges $-q,\, q,\,q$ are placed at the vertices $A$, $B$, $C$ respectively of an equilateral triangle of side $'a'$ as shown in the figure. If charge $-q$ is released keeping remaining two charges fixed, then the kinetic energy of charge $(-q)$ at the instant when it passes through the mid point $M$ of side $BC$ is 

A proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially $\alpha$-particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be