$A$ drop of $10^{-6} \ kg$ water carries $10^{-6} \ C$ charge. What electric field should be applied to balance its weight? (Assume $g = 10 \ m/s^2$)

$A$ uniform electric field of $10\,N/C$ is created between two parallel charged plates (as shown in the figure). An electron enters the field symmetrically between the plates with a kinetic energy of $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of the electron as it comes out of the field is $.........$ (in degrees).

Two charged particles of masses in the ratio $1: 3$ have charges in reciprocal ratio as their masses. They are placed in a uniform electric field and allowed to move. The ratio of their kinetic energies is

An electron and a proton are in a uniform electric field. The ratio of their accelerations will be

$A$ particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is . . . . . . .

$A$ particle of mass $2 \times 10^{-5} \ kg$ and charge $4 \times 10^{-3} \ C$ starts from rest in a uniform electric field of $5 \ V/m$. What is its kinetic energy after $10 \ s$?