The potential difference between two metal plates is $800 \, V$ and they are separated by a horizontal distance of $0.02 \, m$. $A$ particle of mass $1.96 \times 10^{-15} \, kg$ is held in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is .......

An electron with mass $9.1 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is placed in an electric field of $1 \times 10^6 \ V/m$. How much time will it take for the electron to reach a velocity equal to $1/10$th of the speed of light?

$A$ charged particle of mass $0.003 \, g$ is held stationary in space by placing it in a downward electric field of $6 \times 10^4 \, N/C$. The magnitude of the charge is:

$A$ point charge $Q$ is fixed. $A$ small charge $q$ and mass $m$ is given a velocity $v_0$ from infinity at a perpendicular distance $r_0$ as shown. If the distance of closest approach is $r_0/2$,find the value of $q$. [Given $mv_0^2 = \frac{Q^2}{4\pi \epsilon_0 r_0}$]

$A$ point charge $Q$ is placed in a uniform electric field $\vec{E} = E_1 \hat{i} + E_2 \hat{j}$ at position $(a, b)$. Find the work done in moving it to position $(c, d)$.

An inclined plane making an angle of $30^{\circ}$ with the horizontal is placed in a uniform horizontal electric field of $200 \, N/C$ as shown in the figure. $A$ body of mass $1 \, kg$ and charge $5 \, mC$ is allowed to slide down from rest at a height of $1 \, m$. If the coefficient of friction is $0.2$,find the time (in $s$) taken by the body to reach the bottom. $\left[ g = 9.8 \, m/s^2, \sin 30^{\circ} = 0.5, \cos 30^{\circ} = \frac{\sqrt{3}}{2} \approx 0.866 \right]$