An electron is moving with constant velocity along the $x$-axis. If a uniform electric field is applied along the $y$-axis,then its path in the $x-y$ plane will be:

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the 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 $y \times 10^5 \ N/C$. The value of $y$ is

In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

The kinetic energy of an electron accelerated through a potential difference of $100 \ V$ is ........

$A$ positively charged particle moving along the horizontal axis with a certain velocity enters a uniform electric field directed along the vertical axis. Its