$A$ cathode ray tube contains a pair of parallel metal plates $1.0 \, cm$ apart and $3.0 \, cm$ long. $A$ narrow horizontal beam of electrons with a velocity $3 \times 10^7 \, m/s$ passes down the tube midway between the plates. When a potential difference of $550 \, V$ is maintained across the plates,it is found that the electron beam is deflected such that it just strikes the end of one of the plates. The specific charge of the electron in $C/kg$ 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 ]

In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

Kinetic energy of an electron accelerated in a potential difference of $100 \, V$ is

Two balls of charge $q_1$ and $q_2$ initially have a velocity of the same magnitude and direction. After a uniform electric field is applied for a certain time,the direction of the velocity of the first ball changes by $60^{\circ}$,and the velocity magnitude is reduced by half. The direction of the velocity of the second ball changes by $90^{\circ}$. In what proportion will the velocity of the second ball change? Determine the magnitude of the charge-to-mass ratio for the second ball if it is equal to $k_1$ for the first ball. The electrostatic interaction between the balls should be neglected.

$A$ drop of $10^{-6} \ kg$ water carries $10^{-6} \ C$ charge. What electric field should be applied to balance its weight? (Assume $g = 10 \ m/s^2$)