The electrostatic potential inside a charged spherical ball is given by $\phi= a{r^2} + b$ where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is:

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

An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on electron 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 :-

If on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$, then the magnitude of electric field at the origin

The electric potential at a point in free space due to charge $Q$ coulomb is  $V=Q$$ \times {10^{11}}\,V$ . The electric field at that point is