$A$ proton is $1840$ times heavier than an electron. When it is accelerated through a potential difference of $1 \ kV$,what will be its kinetic energy in $keV$?

The coordinates of $A, B, C$ and $D$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$ respectively. How much work is done in moving a charge $q$ from $A$ to $D$ in a uniform electric field $\vec{E}$ directed along the positive $x$-axis?

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 $q = 1.6 \times 10^{-19} \, C$).

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 $2$? $(e = 1.6 \times 10^{-19}\ C, m_0 = 9.11 \times 10^{-31}\ kg)$

$A$ particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

$A$ charged particle of mass $5 \times 10^{-5} \ kg$ is held stationary in space by placing it in an electric field of strength $10^7 \ N C^{-1}$ directed vertically downwards. The charge on the particle is