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 potential at a point $x$ (measured in $\mu m$) due to some charges situated on the $x$-axis is given by $V(x) = \frac{20}{x^2 - 4} \text{ volt}$. The electric field $E$ at $x = 4 \mu m$ is given by:

Two plates are separated by a distance of $20 \, cm$. The potential difference between them is $10 \, V$. The electric field between the two plates is ...... $V m^{-1}$.

In a certain region of space with volume $0.2 \ m^3$,the electric potential is found to be $5 \ V$ throughout. The magnitude of the electric field in this region is . . . . . . $N/C$.

For a charged spherical ball,electrostatic potential inside the ball varies with $r$ as $V = 2ar^2 + b$. Here,$a$ and $b$ are constants 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.

Find the potential $V$ of an electrostatic field $\vec{E} = a(y\hat{i} + x\hat{j})$,where $a$ is a constant.