The electric field in a certain region is acting radially outward and is given by E= Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by:

  • Option 1)

    4\pi \varepsilon _0Aa^{2}

  • Option 2)

    A \varepsilon _0a^{2}

  • Option 3)

    4\pi \varepsilon _0Aa^{3}

  • Option 4)

    \varepsilon _0Aa^{3}

 

Answers (1)

As we discussed 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.

 

Given E=A.r

r=radial\:\: distance

At\:\:\:r=a,\:\:E=Aa

Net flux emitted from spherical surface of radius a is \phi _{net}=\frac{q_{en}}{\epsilon_{o}}

(Aa)\times (4\pi a ^{2})=\frac{q}{\epsilon_{o}}

\therefore q= 4 \pi \epsilon_{o}Aa^{3}


Option 1)

4\pi \varepsilon _0Aa^{2}

Incorrect

Option 2)

A \varepsilon _0a^{2}

Incorrect

Option 3)

4\pi \varepsilon _0Aa^{3}

Correct

Option 4)

\varepsilon _0Aa^{3}

Incorrect

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