$A$ charged oil drop is suspended in a uniform electric field of $3 \times 10^{4} \; V/m$ so that it neither falls nor rises. The charge on the drop will be $..... \times 10^{-18} \; C$. (Take the mass of the drop $= 9.9 \times 10^{-15} \; kg$ and $g = 10 \; m/s^{2}$)

As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

Consider a particle of mass $1 \text{ g}$ and charge $1.0 \text{ C}$ is at rest. Now the particle is subjected to an electric field $E(t) = E_0 \sin(\omega t)$ in the $x$-direction, where $E_0 = 2 \text{ N/C}$ and $\omega = 1000 \text{ rad/s}$. The maximum speed attained by the particle is: (in $\text{ m/s}$)

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

$A$ 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

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?