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Question

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}}.........(

Vedclass · Accepted Answer

$- 6$$a{\varepsilon _0}$

Vedclass · Answer

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, $\rho=\frac{d q}{4 \pi r^{2} d r}=-6 \varepsilon_{0} a$

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:

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