In Millikan's oil drop experiment an oil drop carrying a charge $Q$ is held stationary by a potential difference $2400\,V$ between the plates. To keep a drop of half the radius stationary the potential difference had to be made $600\,V$. What is the charge on the second drop

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

In a region, the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$, where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2$ coulomb situated at point $( 1, 1, 1)$ 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 ($e$) $=1.6 \times 10^{-19} C$ )

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

For a charged spherical ball, electrostatic potential inside the ball varies with $r$ as $V =2 ar ^2+ b$. Here, $a$ and $b$ are constant and $r$ is the distance from the center. The volume charge density inside the ball is $-\lambda a \varepsilon$. The value of $\lambda$ is $...........$. $\varepsilon=$ permittivity of medium.