A charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 × 10^4 $  $Vm^{-1}$. The charge on the drop is $8 × 10^{-18}$ $C$. The radius of the drop is $\left[ {{\rho _{air}} = 1.29\,kg/{m^3};{\rho _{Hg}} = 13.6 \times {{10}^3}kg/{m^3}} \right]$

A thin disc of radius $b = 2a$  has a concentric hole of radius $'a'$ in it (see figure). It carries uniform surface charge $'\sigma '$ on it. If the electric field on its axis at height $'h'$ $(h << a)$ from its centre is given as $'Ch'$ then value of $'C'$ is

For the given figure the direction of electric field at $A$ will be

The acceleration of an electron in an electric field of magnitude $50\, V/cm$, if $e/m$ value of the electron is $1.76 \times {10^{11}}\,C/kg$, is

An oil drop of radius $2\, mm$ with a density $3\, g$ $cm ^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5}\, V\, m ^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider $\left. g =9.81\, m / s ^{2}\right)$

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2} L$.