An electron having charge '$e$' and mass '$m$' is moving in a uniform electric field $E$. Its acceleration will be

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 $\text{ m/s}$)

Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

$A$ 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 $........... \text{ mm}$.

Assertion : $A$ deuteron and an $\alpha -$ particle are placed in an electric field. If $F_1$ and $F_2$ are the forces acting on them and $a_1$ and $a_2$ are their accelerations respectively,then $a_1 = a_2$.
Reason : Forces will be same in an electric field.

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: