Two plates are at potentials $-10\, V$ and $+30\, V$. If the separation between the plates is $2\, cm$,the electric field between them is.......$V/m$.

The electric field vector in a region is given by $E = (3 \hat{i} + 4y \hat{j}) \ V \ m^{-1}$. The potential at the origin is zero. Then,the potential at a point $(2, 1) \ m$ is: (in $V$)

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:

Assume that an electric field $\overrightarrow{E} = 30x^2 \hat{i}$ exists in space. Then the potential difference $V_A - V_O$,where $V_O$ is the potential at the origin and $V_A$ is the potential at $x = 2 \, m$,is: (in $, V$)

$A$ parallel plate capacitor has a potential of $20 \ kV$ and a capacitance of $2 \times 10^{-4} \ \mu F$. If the area of the plate is $0.01 \ m^2$ and the distance between the plates is $2 \ mm$,find the potential gradient.

The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field at that point is: