An electron falls from rest through a distance of $1.5 \text{ cm}$ in a uniform electric field of magnitude $2.0 \times 10^4 \text{ N/C}$. The time taken to cover this distance in seconds is . . . . . . . $(e = 1.6 \times 10^{-19} \text{ C}, m_e = 9.11 \times 10^{-31} \text{ kg})$

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

$A$ small sphere of mass $m = 0.5 \, kg$ carrying a positive charge $q = 110 \, \mu C$ is connected to a light,flexible,and inextensible string of length $r = 60 \, cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 \, N/C$ exists in the space,what is the minimum velocity of the sphere required at the highest point so that it may just complete the circle? $(g = 10 \, m/s^2)$

The $e/m$ ratio for an electron is $1.8 \times 10^{11} \ C \ kg^{-1}$. What will be its velocity when accelerated through a potential difference of $9 \ V$?

$A$ charged particle (mass $m$ and charge $q$) moves along the $X$-axis with velocity $V_{0}$. When it passes through the origin,it enters a region having a uniform electric field $\overrightarrow{E} = -E \hat{j}$ which extends up to $x = d$. What is the equation of the path of the electron in the region $x > d$?

$A$ particle,of mass $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .