An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

$A$ particle having mass $m$ and charge $q$ is at rest. On applying a uniform electric field $E$ on it,it starts moving. When this particle travels a distance $x$ in the direction of the force,its kinetic energy will be . . . . . . .

$A$ mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3\,s$ in this region is.......$m/s$.

In the absence of gravity,a charge $q$ and mass $2m$ is placed stationary in a uniform electric field of intensity $E$. When the charge is released,its speed after $n$ seconds is . . . . . . .

$A$ proton is located at coordinates $(x, y) = (0, 0)$,and an electron is at $(d, h)$,where $d >> h$. At time $t = 0$,a uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible,at what $y$ coordinate will the two particles have the same $y$ position at the same time?

$A$ particle with charge $e$ and mass $m$, moving along the $X$-axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is