$A$ 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:

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,V/m$. At equilibrium,the angle that the pendulum makes with the vertical is (take $g = 10\,m/s^2$)

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?

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be:

$A$ ball of mass $1 \text{ g}$ having a charge of $20 \mu\text{C}$ is tied to one end of a string of length $0.9 \text{ m}$ and can rotate in a vertical plane in a uniform electric field of $100 \text{ NC}^{-1}$ directed upwards. The minimum horizontal velocity that must be given to the ball at the lowest position so that it completes the vertical circle is (Let $g = 10 \text{ ms}^{-2}$): (in $\text{ ms}^{-1}$)

An electron enters an electric field of $9.1 \times 10^6 \ N/C$. The acceleration produced is ..... $m/s^2$.