What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

$A$ 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

$A$ particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0}(1 - ax^{2})$ in the $x$-direction,where $a$ and $E_{0}$ are constants. Initially,the particle was at rest at $x = 0$. Other than the initial position,the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

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

$A$ moving charge will gain energy due to the application of

An electron and a positron enter a uniform electric field $E$ perpendicular to it with equal speeds at the same time. The distance of separation between them in the direction of the field after a time $t$ is (where $\frac{e}{m}$ is the specific charge of the electron).