If the electric flux entering and leaving an enclosed surface respectively is \phi _{1} and \phi _{2} the electric charge inside the surface will be

  • Option 1)

    \left ( \phi _{2}-\phi _{1} \right )\varepsilon _{0}

  • Option 2)

    \left ( \phi _{1}+\phi _{2} \right )/\varepsilon _{0}

  • Option 3)

    \left ( \phi _{2}-\phi _{1} \right )/\varepsilon _{0}

  • Option 4)

    \left ( \phi _{1}+\phi _{2} \right )/\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.

 

 

According to Gauss theorem,

\left ( \phi _{2} -\phi _{1}\right )= \frac{Q}{\varepsilon _{0}}\Rightarrow Q= \left ( \phi _{2} -\phi _{1}\right )\varepsilon _{0}

The flux enters the enclosure if one has a negative charge \left ( -q _{2} \right ) and flux goes out if one has a +ve charge \left ( +q _{1} \right )

As one does not know whether \phi _{1} > \phi _{2}\or \, \, \, \phi _{2} > \phi _{1}, Q=\left ( \phi _{2} -\phi _{1}\right )\varepsilon _{0}


Option 1)

\left ( \phi _{2}-\phi _{1} \right )\varepsilon _{0}

Correct

Option 2)

\left ( \phi _{1}+\phi _{2} \right )/\varepsilon _{0}

Incorrect

Option 3)

\left ( \phi _{2}-\phi _{1} \right )/\varepsilon _{0}

Incorrect

Option 4)

\left ( \phi _{1}+\phi _{2} \right )/\varepsilon _{0}

Incorrect

N neha

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