$A$ drop of $10^{-6} \ kg$ water carries $10^{-6} \ C$ charge. What electric field should be applied to balance its weight? (Assume $g = 10 \ m/s^2$)

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T/S$:

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is: