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

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

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

The electrostatic potential inside a charged spherical ball is given by : $V = b -ar^2$, where  $r$ is the distance from the centre ; $a$ and $b$ are constants. Then, the charge density inside the ball is :

A sphere carrying charge of $Q$ having weight $w$ falls under gravity between a pair of vertical plates at a distance of $d$ from each other. When a potential difference $V$ is applied between the plates the acceleration of sphere changes as shown in the figure, to along line $BC$. The value of $Q$ is :-

Figure shows three points $A$, $B$ and $C$ in a region of uniform electric field $\overrightarrow E $. The line $AB$ is perpendicular and $BC$ is parallel to the field lines. Then which of the following holds good. Where ${V_A} > {V_B}$ and ${V_C}$ represent the electric potential at points $A$, $B$ and $C$ respectively