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)

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.

Qenc - 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

N neha

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