The potential due to an electrostatic charge distribution is $V(r)=\frac{q e^{-\alpha e r}}{4 \pi \varepsilon_{0} r}$, where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/ \alpha$ is

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

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 :

The electric potential in a region is represented as $V = 2x + 3y -z$ ; then the expression of electric field strength is

Within a spherical charge distribution of charge density $\rho \left( r \right)$, $N$ equipotential surfaces of potential ${V_0},{V_0} + \Delta V,{V_0} + 2\Delta V,$$…..{V_0} + N\Delta V\left( {\Delta V > 0} \right),$ are drawn and have increasing radii $r_0, r_1, r_2,……r_N$, respectively. If the difference in the radii of the surfaces is constant for all values of $V_0$ and $\Delta V$ then