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:

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).

An oil drop of radius $2 \, mm$ with a density $3 \, g \, cm^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} \, V \, m^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Consider $g = 9.81 \, m \, s^{-2}$)

In an electron gun,the electrons are accelerated by the potential $V$. If $e$ is the charge and $m$ is the mass of an electron,then the maximum velocity of these electrons will be

Consider a particle of mass $1 \text{ g}$ and charge $1.0 \text{ C}$ is at rest. Now the particle is subjected to an electric field $E(t) = E_0 \sin(\omega t)$ in the $x$-direction, where $E_0 = 2 \text{ N/C}$ and $\omega = 1000 \text{ rad/s}$. The maximum speed attained by the particle is: (in $\text{ m/s}$)

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