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 $g = 10 \ m/s^2$.)

An electron is accelerated through a potential difference $(p.d.)$ of $45.5 \ V$. The velocity acquired by it is (in $m \ s^{-1}$)

$A$ simple pendulum of length $L$ is placed between the plates of a parallel plate capacitor having an electric field $E,$ as shown in the figure. Its bob has mass $m$ and charge $q.$ The time period of the pendulum is given by:

$A$ particle of mass $m$ and charge $q$ is fastened to one end $A$ of a massless string having equilibrium length $\ell$,whose other end is fixed at point $O$. The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field $E$ is switched on along the $x$-direction as shown in the figure,then the speed of the particle when it crosses the $x$-axis is

$A$ proton and an $\alpha$-particle are placed in a uniform electric field. The accelerations produced in them are $a_p$ and $a_\alpha$ respectively. Then $a_p : a_\alpha$ will be . . . . . . .

$A$ particle of mass $2 \times 10^{-5} \ kg$ and charge $4 \times 10^{-3} \ C$ starts from rest in a uniform electric field of $5 \ V/m$. What is its kinetic energy after $10 \ s$?