In Millikan's oil drop experiment,an oil drop of mass $16 \times 10^{-6} \ kg$ is balanced by an electric field of $10^6 \ V/m$. The charge in coulomb on the drop,assuming $g = 10 \ m/s^2$,is:

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:

The tiny ball at the end of the thread shown in the figure has a mass of $0.5 \, g$ and is placed in a horizontal electric field of intensity $500 \, N/C$. It is in equilibrium in the position shown. The magnitude and sign of the charge on the ball is .....$\mu C$.

$A$ 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

$A$ simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having an electric field $E,$ as shown in the figure. Its bob has mass $m$ and charge $q.$ The time period of the pendulum is given by:

$A$ uniform electric field $\vec{E}$ exists between the plates of a charged condenser. $A$ charged particle enters the space between the plates perpendicular to $\vec{E}$. The path of the particle between the plates is a