In a Millikan oil drop experiment,a charged drop of mass $1.8 \times 10^{-14} \ kg$ is stationary between its plates. The distance between the plates is $0.90 \ cm$ and the potential difference is $2.0 \ kV$. The number of electrons on the drop is:

An electron enters an electric field of $9.1 \times 10^6 \ N/C$. The acceleration produced is ..... $m/s^2$.

$A$ $2 \ g$ object,located in a region of uniform electric field $E = (300 \ NC^{-1}) \hat{i}$,carries a charge $Q$. The object is released from rest at $x = 0$ and has a kinetic energy of $0.12 \ J$ at $x = 0.5 \ m$. Then,$Q$ is: (in $\mu C$)

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

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

$A$ particle of mass $m$ and charge $q$ travelling with a velocity $v$ along the $x$-axis enters a uniform electric field $\overrightarrow{E}$ directed along the $y$-axis. What will be the trajectory of the particle?