An electron is accelerated through a potential difference of $1000 \ V$. Its velocity is nearly

$A$ particle of mass $M$ and charge $q$,initially at rest,is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)

$A$ particle of mass $1 \, g$ and charge $-0.1 \, \mu C$ is projected from the ground with a velocity $10\sqrt{2} \, m/s$ at an angle of $45^o$ with the horizontal in a region having a uniform electric field $1 \, kV/cm$ in the horizontal direction. Acceleration due to gravity is $10 \, m/s^2$ in the vertical downward direction. Select the $INCORRECT$ statement.

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.

In Millikan's oil drop experiment,an oil drop of mass $16 \times 10^{-6} \ kg$ is balanced by an electric field of $10^6 \ V/m$. The charge in coulomb on the drop,assuming $g = 10 \ m/s^2$,is: