Electrostatic potential inside a charged spherical ball is given by phi= a{r^2} + b where r is dista

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2. Electric Potential and Capacitance

hard

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:

A

$-24\pi a{\varepsilon _0}r$

B

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

C

$-24$$\pi a{\varepsilon _0}$

D

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

(AIEEE-2011)

Solution

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

Standard 12

Physics

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