The electric potential $V$ at any point $O$ ($x$, $y$, $z$ all in metres) in space is given by $V = 4{x^2}\,volt$. The electric field at the point $(1m,\,0,\,2m)$ in $volt/metre$ 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 ($e$) $=1.6 \times 10^{-19} C$ )

The electrostatic potential inside a charged spherical ball is given by $\phi= a{r^2} + b$ where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is:

Two plates are $2\,cm$ apart, a potential difference of $10\;volt$ is applied between them, the electric field between the plates is………$N/C$

A spherical charged conductor has surface charge density $\sigma $ . The electric field on its surface is $E$ and electric potential of conductor is $V$ . Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be