The electric field at a point is given by $\vec E = E_0 \hat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path as shown in the figure. Find the work done by the electric field in this motion.

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on the electron 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?

$A$ horizontal electric field $E = (mg)/q$ exists as shown in the figure. $A$ mass $m$ with charge $q$ is attached to the end of a light rod of length $l$. If the mass $m$ is released from the position shown in the figure $(\theta = 45^{\circ})$,find the angular velocity of the rod when it passes through the bottom-most position.

An electron moving with a speed of $5 \times 10^6 \, m/s$ is shot parallel to an electric field of intensity $1 \times 10^3 \, N/C$. The field causes retardation of the electron's motion. Evaluate the distance traveled by the electron before coming to rest for an instant (mass of $e = 9 \times 10^{-31} \, kg$,charge $q = 1.6 \times 10^{-19} \, C$).