A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ in the $x-$direction, where $a$ and $E _{0}$ are constants. Initially the particle was at rest at $x=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
$\sqrt{\frac{2}{a}}$
$\sqrt{\frac{1}{a}}$
$a$
$\sqrt{\frac{3}{a}}$
Charge $q_{2}$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_{1}$, once around this equipotential path, would be
Four identical charges $ + \,50\,\mu C$ each are placed, one at each corner of a square of side $2\,m$. How much external energy is required to bring another charge of $ + \,50\,\mu C$ from infinity to the centre of the square......$J$ $\left( {{\rm{Given}}\frac{{\rm{1}}}{{{\rm{4}}\pi {\varepsilon _{\rm{0}}}}} = 9 \times {{10}^9}\,\frac{{N{m^2}}}{{{C^2}}}} \right)$
Nine point charges are placed on a cube as shown in the figure. The charge $q$ is placed at the body centre whereas all other charges are at the vertices. The electrostatic potential energy of the system will be
For equal point charges $Q$ each are placed in the $xy$ plane at $(0, 2), (4, 2), (4, -2)$ and $(0, -2)$. The work required to put a fifth change $Q$ at the origin of the coordinate system will be
Two positrons $(e^+)$ and two protons $(p)$ are kept on four corners of a square of side $a$ as shown in figure. The mass of proton is much larger than the mass of positron. Let $q$ denotes the charge on the proton as well as the positron then the kinetic energies of one of the positrons and one of the protons respectively after a very long time will be-