$A$ charged oil drop is suspended in a uniform electric field of $3 \times 10^{4} \; V/m$ so that it neither falls nor rises. The charge on the drop will be $..... \times 10^{-18} \; C$. (Take the mass of the drop $= 9.9 \times 10^{-15} \; kg$ and $g = 10 \; m/s^{2}$)

An electric field due to a positively charged long straight wire at a distance $r$ from it is proportional to $r^{-1}$ in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii $1 \mathring{A}$ and $2 \mathring{A}$. The ratio of their respective time periods is

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 $\text{ m/s}$)

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

$A$ uniform vertical electric field $E$ is established in the space between two large parallel plates. $A$ small conducting sphere of mass $m$ is suspended in the field from a string of length $L$. If the sphere is given a $+q$ charge and the lower plate is charged positively,the period of oscillation of this pendulum is:

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: