An electron falls from rest through a distance of $1.5 \text{ cm}$ in a uniform electric field of magnitude $2.0 \times 10^4 \text{ N/C}$. The time taken to cover this distance in seconds is . . . . . . . $(e = 1.6 \times 10^{-19} \text{ C}, m_e = 9.11 \times 10^{-31} \text{ kg})$

$A$ particle of mass $m$ and charge $q$ is released from rest in a region of uniform electric field of intensity $E$. After traveling a distance $y$,the kinetic energy of the particle is:

$A$ block of mass $m$ containing a net negative charge $-q$ is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant $k$ as shown. If a horizontal electric field $E$ parallel to the spring is switched on,then the maximum compression of the spring is:

In a Millikan oil drop experiment,a charged drop of mass $1.8 \times 10^{-14} \ kg$ is stationary between its plates. The distance between the plates is $0.90 \ cm$ and the potential difference is $2.0 \ kV$. The number of electrons on the drop is:

Two large parallel plates are charged such that the potential difference between them is $V_2 - V_1 = 20\ V$. Plate $2$ is at a higher potential. The plates are separated by a distance of $0.1\ m$ and can be considered infinitely large. An electron is released from rest at the inner surface of plate $1$. What is its speed when it strikes plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_0 = 9.11 \times 10^{-31}\ kg)$

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)