$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

In a parallel plate air capacitor,a cathode beam comprising $n = 10^6$ electrons is emitted with a velocity $v_0 = 10^8\ m/s$ into the space between the plates. The potential difference between the plates is $\phi = 400\ V$,the separation between the plates is $d = 2\ cm$ and the area of each plate is $l^2 = 100\ cm^2$ (where $l$ is the length of the plate). The deflection of the electron beam is ........... $mm$.

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the electric field that is needed if the electron is to emerge from an exit hole $1.0 \ cm$ away from the entrance hole,traveling at right angles to its original direction,is $y \times 10^5 \ N/C$. The value of $y$ is

$A$ liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5 \times 10^3 \, Vm^{-1}$. The density of the liquid is $1.26 \times 10^3 \, kg \, m^{-3}$. The radius of the drop is (neglect buoyancy):

$A$ charged oil drop is suspended in a uniform electric field of $3 \times 10^{4} \; V/m$ so that it neither falls nor rises. The charge on the drop will be $..... \times 10^{-18} \; C$. (Take the mass of the drop $= 9.9 \times 10^{-15} \; kg$ and $g = 10 \; m/s^{2}$)

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is: