$A$ small sphere carrying a charge $q$ is hanging in between two parallel plates by a string of length $L$. The time period of the pendulum is $T_0$. When the parallel plates are charged,the time period changes to $T$. The ratio $T/T_0$ is equal to

An electron falls through a distance of $1.5 \; cm$ in a uniform electric field of magnitude $2.0 \times 10^{4} \; N C^{-1}$ [Figure $(a)$]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Figure $(b)$]. Compute the time of fall in each case. Contrast the situation with that of 'free fall under gravity'.

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 10^3 \ Vm^{-1}$ acting vertically upwards as shown in the figure. The plates are sufficiently long and have a separation of $2 \ cm$. $A$ particle of negative charge $1 \ \mu C$ and mass $2 \ g$ is projected at an angle $45^{\circ}$ with the electric field from the lower plate with a velocity '$u$'. The maximum velocity acquired by the particle,if it does not hit the upper plate,is (in $ms^{-1}$)

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

$A$ particle,of mass $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.