Consider a particle of mass $1 \text{ g}$ and charge $1.0 \text{ C}$ is at rest. Now the particle is subjected to an electric field $E(t) = E_0 \sin(\omega t)$ in the $x$-direction,where $E_0 = 2 \text{ N/C}$ and $\omega = 1000 \text{ rad/s}$. The maximum speed attained by the particle is: (in $\text{ m/s}$)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.

$A$ liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5 \times 10^3 \, Vm^{-1}$. The density of the liquid is $1.26 \times 10^3 \, kg \, m^{-3}$. The radius of the drop is (neglect buoyancy):

$A$ stream of positively charged particles having $\frac{q}{m} = 2 \times 10^{11} \text{ C/kg}$ and velocity $\overrightarrow{v}_0 = 3 \times 10^7 \hat{i} \text{ m/s}$ is deflected by an electric field $1.8 \hat{j} \text{ kV/m}$. The electric field exists in a region of $10 \text{ cm}$ along the $x$-direction. Due to the electric field,the deflection of the charged particles in the $y$-direction is $........... \text{ mm}$.

In Millikan's oil drop experiment,an oil drop of mass $16 \times 10^{-6} \ kg$ is balanced by an electric field of $10^6 \ V/m$. The charge in coulomb on the drop,assuming $g = 10 \ m/s^2$,is: