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

In a uniform electric field of $E = 1 \times 10^{4} \text{ N C}^{-1}$,an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \text{ m}$ is nearly (given $\frac{e}{m} \text{ of electron } \approx 1.8 \times 10^{11} \text{ C kg}^{-1}$):

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with speed $v_{x}$ (like particle $1$ in the Figure). The length of the plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $q E L^{2} / (2 m v_{x}^{2})$. Compare this motion with the motion of a projectile in a gravitational field.

$A$ moving charge will gain energy due to the application of

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 $ms^{-1}$)

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