Two charged particles of masses $m$ and $2m$ have charges $+2q$ and $+q$ respectively. They are kept in a uniform electric field and allowed to move for the same time. The ratio of their kinetic energies will be

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (Given: mass of an electron $m = 9.1 \times 10^{-31} \ kg$,charge of an electron $e = 1.6 \times 10^{-19} \ C$,and acceleration due to gravity $g = 10 \ m/s^2$.)

$A$ particle of mass $0.2 \ g$ and charge $2 \ C$ is released from rest in a uniform electric field of $20 \ N \ C^{-1}$. The kinetic energy of the particle after moving a distance of $20 \ cm$ is (in $J$)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

$A$ uniform vertical electric field $E$ is established in the space between two large parallel plates. $A$ small conducting sphere of mass $m$ is suspended in the field from a string of length $L$. If the sphere is given a $+q$ charge and the lower plate is charged positively,the period of oscillation of this pendulum is:

The intensity of the electric field required to keep a water drop of radius $10^{-5} \, cm$ and carrying a charge of $1$ electron suspended in air is: