$A$ 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:

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\ V$ (i.e.,plate $2$ is at a higher potential). The plates are separated by $d = 0.1\ m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1$. What is its speed when it hits plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_e = 9.11 \times 10^{-31}\ kg)$

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

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 $ms^{-1}$)

$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

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'.