$A$ small oil drop of mass $10^{-6} \,kg$ is hanging at rest between two plates separated by $1 \,mm$ having a potential difference of $500 \,V$. The charge on the drop is $(g = 10 \,ms^{-2})$

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,V/m$. At equilibrium,the angle that the pendulum makes with the vertical is (take $g = 10\,m/s^2$)

$A$ water drop of mass $10^{-6} \ kg$ carries a charge of $(-10^{-6}) \ C$. What magnitude of electric field should be applied to the drop so that it is in equilibrium with its weight?

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

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.