$A$ uniform electric field of $10^3 \, V/m$ is directed along the $y$-axis. $A$ particle of mass $1 \, g$ and charge $10^{-6} \, C$ is projected from the origin into the field with a velocity of $10 \, m/s$ along the positive $x$-axis. Its speed after $10 \, s$ is (in $m/s$):

An electron moving with a speed of $5 \times 10^6 \, m/s$ is shot parallel to an electric field of intensity $1 \times 10^3 \, N/C$. The field causes retardation of the electron's motion. Evaluate the distance traveled by the electron before coming to rest for an instant (mass of $e = 9 \times 10^{-31} \, kg$,charge $q = 1.6 \times 10^{-19} \, C$).

$A$ charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If the charge on it is equal to the charge of an electron,then the intensity of the electric field will be.......$N/C$ $(g = 10\,m/s^2)$

As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

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

$A$ positive charge particle of $100 \,mg$ is thrown in the opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms^{-1}$,how much distance (in $m$) will it travel before coming to rest momentarily?