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

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

$A$ vertical electric field of magnitude $4.9 \times 10^{5} \, N/C$ just prevents a water droplet of mass $0.1 \, g$ from falling. The value of charge on the droplet will be ........ $\times 10^{-9} \, C$ (Given $g = 9.8 \, m/s^{2}$)

An electron enters an electric field with its velocity in the direction of the electric lines of force. Then:

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'.

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be: