$A$ charged particle is free to move in an electric field. It will travel

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the final tension in the string changes. The final tension in the string, when the pendulum attains an equilibrium position, is . . . . . . . ($g$: acceleration due to gravity)

An electron is rotating around an infinite positive linear charge in a circle of radius $0.1 \,m$. If the linear charge density is $1 \,\mu C/m$,then the velocity of the electron in $m/s$ will be ...... $\times 10^7$.

$A$ charge of magnitude $2e$ and mass $4m$ is moving in an electric field $E$. The acceleration imparted to the above charge is

$A$ 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

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?