The acceleration of an electron in an electric field of magnitude $50\, V/cm$ is (given that the $e/m$ ratio of the electron is $1.76 \times 10^{11}\, C/kg$):

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

$A$ 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

An electron is accelerated through a potential difference of $200 \ V$. If $e/m$ for the electron is $1.6 \times 10^{11} \ C/kg$,the velocity acquired by the electron will be:

$A$ charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 \times 10^4 \ Vm^{-1}$. The charge on the drop is $8 \times 10^{-18} \ C$. The radius of the drop is $\left[ \rho_{air} = 1.29 \ kg/m^3, \rho_{Hg} = 13.6 \times 10^3 \ kg/m^3, g = 9.8 \ m/s^2 \right]$.

$A$ horizontal electric field $E = (mg)/q$ exists as shown in the figure. $A$ mass $m$ with charge $q$ is attached to the end of a light rod of length $l$. If the mass $m$ is released from the position shown in the figure $(\theta = 45^{\circ})$,find the angular velocity of the rod when it passes through the bottom-most position.