An electron falls through a distance of $1.5 \, cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \, N/C$ as shown in the figure. The time taken by the electron to fall through this distance is ($m_e = 9.1 \times 10^{-31} \, kg$,neglect gravity).

The electric potential varies in space according to the relation $V = 3x + 4y$. $A$ particle of mass $0.1 \, kg$ starts from rest from point $(2, 3.2)$ under the influence of this field. The charge on the particle is $+1 \, \mu C$. Assume $V$ and $(x, y)$ are in $S.I.$ units. The time taken to cross the $x$-axis is.....$s$

$A$ proton is fired at an initial velocity of $150 \, m/s$ at an angle of $60^\circ$ above the horizontal into a uniform electric field of $2 \times 10^{-4} \, N/C$ between two charged parallel plates as shown in the figure. The total time the particle is in motion is:

$A$ 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

$A$ 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 with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be: