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

Option 1)
$-24\pi a\varepsilon _{0}r$

Option 2)
$-6 a\varepsilon _{0}r$

Option 3)
$-24\pi a\varepsilon _{0}$

Option 4)
$-6 a\varepsilon _{0}$

Answers (2)

best_answer

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

$\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}$

- wherein

No need to be a real physical surface.

Q_enc - charge enclosed by closed surface.

$\phi=ar^{2}+b$

Electric field $E=\frac{-d\phi}{dr}=-2ar$ - (i)

$\phi=\int \vec{E}.d\vec{S}=\frac{q}{\varepsilon _{0}}$

$-2ar\:4\pi r^{2}=\frac{q_{in}}{\epsilon _{0}}$

$q_{inside}=-8\varepsilon _{0}a\pi\:r^{3}$

change density inside the ball is

$\rho _{inside}=\frac{q_{inside}}{\tfrac{4}{3}\pi\:r^{3}}$

$=\frac{-8\varepsilon _{0}a\pi\:r^{3}}{\tfrac{4}{3}\pi\:r^{3}}$

$\rho _{inside}=-6a\varepsilon _{0}$

Option 1)

$-24\pi a\varepsilon _{0}r$

This is incorrect option

Option 2)

$-6 a\varepsilon _{0}r$

This is incorrect option

Option 3)

$-24\pi a\varepsilon _{0}$

This is incorrect option

Option 4)

$-6 a\varepsilon _{0}$

This is correct option

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prateek

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