An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is at $(d, h)$,where $d >> h$. At time $t = 0$,a uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible,at what $y$ coordinate will the two particles have the same $y$ position at the same time?

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

Two large parallel plates are charged such that the potential difference between them is $V_2 - V_1 = 20\ V$. Plate $2$ is at a higher potential. The plates are separated by a distance of $0.1\ m$ and can be considered infinitely large. An electron is released from rest at the inner surface of plate $1$. What is its speed when it strikes plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_0 = 9.11 \times 10^{-31}\ kg)$

In a region of space,suppose there exists a uniform electric field $\vec{E} = 10 \hat{i} \text{ V/m}$. If a positive charge moves with a velocity $\vec{v} = -2 \hat{j} \text{ m/s}$,its potential energy:

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