An electron having charge ‘$e$’ and mass ‘$m$’ is moving in a uniform electric field $E$. Its acceleration will be
$\frac{{{e^2}}}{m}$
$\frac{{{E^2}e}}{m}$
$\frac{{eE}}{m}$
$\frac{{mE}}{e}$
A proton and an $\alpha$-particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure
A particle of charge $1\ \mu C\ \&\ mass$ $1\ gm$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ sec$. Then it's final speed cannot be.......$m/s$
There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)
An electron enters a parallel plate capacitor with horizontal speed $u$ and is found to deflect by angle $\theta$ on leaving the capacitor as shown below. It is found that $\tan \theta=0.4$ and gravity is negligible. If the initial horizontal speed is doubled, then the value of $\tan \theta$ will be
A proton sits at coordinates $(x, y) = (0, 0)$, and an electron 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, the $y$ coordinates of the two particles will be equal (at equal time)