$A$ charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If the charge on it is equal to the charge of an electron,then the intensity of the electric field will be.......$N/C$ $(g = 10\,m/s^2)$

If a proton is moved against the Coulomb force of an electric field,then

An electron is accelerated through a potential difference of $1000 \ V$. Its velocity is nearly

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

Two equal negative charges $-q$ are fixed at the points $(0, a)$ and $(0, -a)$ on the $y-$ axis. $A$ positive charge $Q$ is released from rest at the point $(2a, 0)$ on the $x-$ axis. The charge will

$A$ 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 $\text{ ms}^{-1}$)