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

$A, B$ and $C$ are three points in a uniform electric field. The electric potential is

Two large circular discs separated by a distance of $0.01 m$ are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900 kg m ^{-3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of $200 V$ across the discs. As a result, an oil drop of radius $8 \times 10^{-7} m$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is (neglect the buoyancy force, take acceleration due to gravity $=10 ms ^{-2}$ and charge on an electron ($e$) $=1.6 \times 10^{-19} C$ )

Figure shows two equipotential lines in $x, y$ plane for an electric field. The scales are marked. The $x-$ component $E_x$ and $y$ -component $E_y$ of the electric field in the space between these equipotential lines are respectively :-

The maximum electric field that can be held in air without producing ionisation of air is $10^7\,V/m$. The maximum potential therefore, to which a conducting sphere of radius $0.10\,m$ can be charged in air is

The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5{x^2} + 10x - 9)\,volt$. Value of electric field at $x = 1$ is......$V/m$