$A$ proton is $1840$ times heavier than an electron. When it is accelerated through a potential difference of $1 \ kV$,what will be its kinetic energy in $keV$?

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

An electron with mass $9.1 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is placed in an electric field of $1 \times 10^6 \ V/m$. How much time will it take for the electron to reach a velocity equal to $1/10$th of the speed of light?

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

$A$ charged cork ball having mass $1 \text{ g}$ and charge $q$ is suspended on a light string in a uniform electric field as shown in the figure. The ball is in equilibrium at $\theta=37^{\circ}$,when the value of the electric field is $E=(3 \hat{i}+5 \hat{j}) \times 10^5 \text{ NC}^{-1}$. (Assume $T$ as tension in the string.) Which of the following options are correct? (Given,$\sin 37^{\circ}=0.60$ and $g=10 \text{ ms}^{-2}$)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$