Electric potential at any point is $V = - 5x + 3y + \sqrt {15} z$, then the magnitude of the electric field is

The electric potential at a point in free space due to charge $Q$ coulomb is  $V=Q$$ \times {10^{11}}\,V$ . The electric field at that point is 

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

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 ($e$) $=1.6 \times 10^{-19} C$ )

If potential (in volts) in a region is expressed as $V (x,y,z) =6xy-y+2yz $ the electric field (in $N/C$) at point $(1, 1, 0)$ is 

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