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Q.

The electrostatic potential inside a charged spherical ball is given by $ϕ = a r^{2} + b$ 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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a

$- 24 π a \in_{o} r$

b

$- 6 a \in_{o} r$

c

$- 24 π a \in_{o}$

d

$- 6 a \in_{o}$

answer is D.

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Detailed Solution

$ϕ = a r^{2} + b$

$E = - \frac{d ϕ}{d t} = - 2 a r$

$\oint \vec{E} . \vec{d s} = \frac{q}{\in_{o}}$

$- 2 a r .4 π r^{2} = \frac{q}{\in_{o}}$

$q = - 8 \in_{o} a π r^{3}$

$P = \frac{q}{\frac{4}{3} π r^{3}}$

$\Rightarrow$ $P = - 6 a \in_{o}$

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The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b where ‘r’ is the distance from the centre and a, b are constants. Then the charge density inside the ball is