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  • Confused! kindly explain, A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

  • Option 1)

    increase four times

  • Option 2)

    be reduced to half

  • Option 3)

    remain the same

  • Option 4)

    be doubled

 

Answers (1)


@9715

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.

 

    electric flux through a surface = \frac{q_{in}}{\epsilon _{0}}

since q_{in} is constant, flux will remain same.

 


Option 1)

increase four times

Incorrect

Option 2)

be reduced to half

Incorrect

Option 3)

remain the same

Correct

Option 4)

be doubled

Incorrect

Posted by

Vakul

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