$A$ vertical electric field of magnitude $4.9 \times 10^{5} \, N/C$ just prevents a water droplet of mass $0.1 \, g$ from falling. The value of charge on the droplet will be ........ $\times 10^{-9} \, C$ (Given $g = 9.8 \, m/s^{2}$)

$A$ charged particle of mass $m$ and charge $q$ is at rest. It is accelerated in a uniform electric field of intensity $E$ for time $t$. The kinetic energy of the particle after time $t$ is

An electron and a positron enter a uniform electric field $E$ perpendicular to it with equal speeds at the same time. The distance of separation between them in the direction of the field after a time $t$ is (where $\frac{e}{m}$ is the specific charge of the electron).

$A$ particle,of mass $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .

$A$ moving charge will gain energy due to the application of

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the electric field that is needed if the electron is to emerge from an exit hole $1.0 \ cm$ away from the entrance hole,traveling at right angles to its original direction,is $y \times 10^5 \ N/C$. The value of $y$ is