In a region of uniform electric field of intensity $E$,an electron of mass $m_e$ is released from rest. The distance travelled by the electron in a time $t$ is

$A$ 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 '$q$' is [ $\epsilon_0$ = permittivity of free space ]

If a proton is moved against the Coulomb force of an electric field,then

$A$ positive charge particle of $100 \,mg$ is thrown in the opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms^{-1}$,how much distance (in $m$) will it travel before coming to rest momentarily?

$A$ 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.

$A$ uniform electric field $\vec{E}$ exists between the plates of a charged condenser. $A$ charged particle enters the space between the plates perpendicular to $\vec{E}$. The path of the particle between the plates is a