$A$ proton and an $\alpha$-particle are placed in a uniform electric field. The accelerations produced in them are $a_p$ and $a_\alpha$ respectively. Then $a_p : a_\alpha$ 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

$A$ hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \,V$. The speed of the electron when it strikes the anode is

$A$ positive charge particle of $100 \,mg$ is thrown in the opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms^{-1}$,how much distance (in $m$) will it travel before coming to rest momentarily?

Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is

$A$ particle with charge $e$ and mass $m$, moving along the $X$-axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is