$A$ proton and an $\alpha$-particle are both accelerated from rest in a uniform electric field. The ratio of work done by the electric field on the proton and the $\alpha$-particle in a given time is

$A$ proton and an $\alpha$-particle having equal kinetic energy are projected into a uniform transverse electric field as shown in the figure.

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

$A$ 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 . . . . . .

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ as shown in the figure. $A$ body of mass $1 \, kg$ and charge $5 \, mC$ is allowed to slide down from rest at a height of $1 \, m$. If the coefficient of friction is $0.2$,find the time (in $s$) taken by the body to reach the bottom. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)