An oil drop having charge $2e$ is kept stationary between two parallel horizontal plates $2.0 \, cm$ apart when a potential difference of $12000 \, V$ is applied between them. If the density of oil is $900 \, kg/m^3$,the radius of the drop will be

$A$ 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

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).

An electron is moving in an electric field and a magnetic field. It will gain energy from:

The tiny ball at the end of the thread shown in the figure has a mass of $0.5 \, g$ and is placed in a horizontal electric field of intensity $500 \, N/C$. It is in equilibrium in the position shown. The magnitude and sign of the charge on the ball is .....$\mu C$.

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is: