In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is

$A$ particle of mass $0.5 \ g$ and charge $10 \ \mu C$ is subjected to a uniform electric field of $8 \ NC^{-1}$. If the particle is initially at rest,the velocity of the particle after a time of $5 \ s$ is: (in $ms^{-1}$)

$A$ 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 $(g = 10 \,ms^{-2})$

An electron and a proton are placed in the same electric field. The ratio of their accelerations is ......

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 charge $q$ and mass $m$ is subjected to an electric field $E = E_{0}(1 - ax^{2})$ in the $x$-direction,where $a$ and $E_{0}$ are constants. Initially,the particle was at rest at $x = 0$. Other than the initial position,the kinetic energy of the particle becomes zero when the distance of the particle from the origin is: