An electron is moving in an electric field and a magnetic field. It will gain energy from:

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\ V$ (i.e.,plate $2$ is at a higher potential). The plates are separated by $d = 0.1\ m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1$. What is its speed when it hits plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_e = 9.11 \times 10^{-31}\ kg)$

In the following diagram,which particle has the highest $e/m$ value?

$A$ particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

An electron falls through a distance of $1.5 \; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \; N C^{-1}$ [Figure $(a)$]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)$]. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).