$A$ 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 $g = 9.8 \, m/s^{2}$)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.

The intensity of the electric field required to keep a water drop of radius $10^{-5} \, cm$ and carrying a charge of $1$ electron suspended in air is:

An electron is accelerated through a potential difference $(p.d.)$ of $45.5 \ V$. The velocity acquired by it is (in $m \ s^{-1}$)

In a uniform electric field,if a charge is fired in a direction different from the line of the electric field,then the trajectory of the charge will be a

In an ink-jet printer,an ink droplet of mass $m$ is given a negative charge $q$ by a computer-controlled charging unit,and then enters at speed $v$ in the region between two deflecting parallel plates of length $L$ separated by distance $d$ (see figure below). All over this region exists a downward electric field $E$ which you can assume to be uniform. Neglecting the gravitational force on the droplet,the maximum charge that can be given so that it will not hit a plate is close to :