A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
$4$
$8$
$10$
$1$
Derive the formula for the electric potential energy of system of three charges.
A charged particle of charge $Q $ is held fixed and another charged particle of mass $m$ and charge $q$ (of the same sign) is released from a distance $r.$ The impulse of the force exerted by the external agent on the fixed charge by the time distance between $Q$ and $q$ becomes $2r$ is
A point charge $q$ is surrounded by eight identical charges at distance $r$ as shown in figure. How much work is done by the forces of electrostatic repulsion when the point charge at the centre is removed to infinity?
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......$J$ $\left( {{\rm{Given}}\frac{{\rm{1}}}{{{\rm{4}}\pi {\varepsilon _{\rm{0}}}}} = 9 \times {{10}^9}\,\frac{{N{m^2}}}{{{C^2}}}} \right)$
In the electric field of a point charge $q$, a certain charge is carried from point $A$ to $B$, $C$, $D$ and $E$. Then the work done