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:
AIEEE 2011, Diffcult
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Electric field
$E=-\frac{d \phi}{d r}=-2 a r.........(i)$
By Gauss's theorem
$E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}.........(ii)$
From $(i)$ and $(ii)$,
$\mathrm{q}=-8 \pi \varepsilon_{0} a r^{3}$
$\Rightarrow \quad d q=-24 \pi \varepsilon_{0} a r^{2} d r$
Charge density, $\rho=\frac{d q}{4 \pi r^{2} d r}=-6 \varepsilon_{0} a$
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