Four closed surfaces and corresponding charge distributions are shown below.
 

Let the respective electric fluxes through the surfaces be Φ1, Φ2, Φ3 and Φ4.  Then :

  • Option 1)

    Φ1 < Φ2=Φ3 > Φ4

  • Option 2)

    Φ1 > Φ2 > Φ3 > Φ4

  • Option 3)

     Φ1=Φ2=Φ3=Φ4

  • Option 4)

    Φ1 > Φ3 ; Φ2 < Φ4

     

 

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 = \frac{q_{enclosed}}{E_{0}}

For\, S_{1}\,: \phi _{1}= \frac{2q}{\epsilon_{0}}\, \, \,\, \, \, \, \, \, For\, S_{2}\,:\, \phi _{2}= \frac{3q-q}{\epsilon_{0}}= \frac{2q}{\epsilon_{0}} \, \,\, \, \, \, \, \,

For\, S_{3}\,: \phi _{3}= \frac{q+q}{\epsilon_{0}}= \frac{2q}{\epsilon_{0}} \, \,\, \, \, \, \, \,For\, S_{4}\,: \phi _{4}= \frac{8q-6q}{\epsilon_{0}}= \frac{2q}{\epsilon_{0}}

Hence \phi _{1}= \phi _{2}= \phi _{3}= \phi _{4}=\frac{2q}{\epsilon_{0}}

 


Option 1)

Φ1 < Φ2=Φ3 > Φ4

Incorrect Option

Option 2)

Φ1 > Φ2 > Φ3 > Φ4

Incorrect Option

Option 3)

 Φ1=Φ2=Φ3=Φ4

Correct Option

Option 4)

Φ1 > Φ3 ; Φ2 < Φ4

 

Incorrect Option

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