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  • Stuck here, help me understand: - Electrostatics - JEE Main-5

The inward and outward electric flux for a closed surface in units of N m^2/C  are respectively 8\times 10^3 and 4\times 10^3 Then the total charge inside the surface is [where \varepsilon _0=  permittivity constant]

  • Option 1)

    4\times 10^3 C

  • Option 2)

    -4\times 10^3 C

  • Option 3)

    -4\times 10^3 C/\varepsilon

  • Option 4)

    4\times 10^3 \varepsilon _0C

 

Answers (1)

best_answer

As we have learned

if Electric field is variable -

\phi = \int \vec{E}\cdot d\vec{A}

-

 

 

By Gauss’s law \phi = 1/\varepsilon _0  (Qenclosed)                     15914

ÞQ_{enclosed}= \phi \varepsilon _0= (-8*10^3+4*10^3)\varepsilon _0


Option 1)

4\times 10^3 C

Option 2)

-4\times 10^3 C

Option 3)

-4\times 10^3 C/\varepsilon

Option 4)

4\times 10^3 \varepsilon _0C

Posted by

Plabita

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