An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

The intensity of the electric field required to keep a water drop of radius $10^{-5} \, cm$ and carrying a charge of $1$ electron suspended in air is:

$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is:

An electron is accelerated through a potential difference of $200 \ V$. If $e/m$ for the electron is $1.6 \times 10^{11} \ C/kg$,the velocity acquired by the electron will be:

$A$ 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? $(g = 10 \, m/s^2)$

If a proton is moved against the Coulomb force of an electric field,then