What is the flux through a cube of side 'a' if a point charge of 'q' is at one of its corner?

  • Option 1)

    \frac{2q}{\varepsilon _0}

  • Option 2)

    \frac{q}{8\varepsilon _0}

  • Option 3)

    \frac{q}{\varepsilon _0}

  • Option 4)

    \frac{q}{2\varepsilon _0}

 

Answers (1)
D Divya Saini

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.

 

 

 

Eight identical cubes are required so that the given charge q appears at the centre of the bigger cube.

\therefore \phi = \frac{1}{8}\left \lfloor \frac{q}{\varepsilon _{0}} \right \rfloor =\frac{q}{8\varepsilon _{0}}


Option 1)

\frac{2q}{\varepsilon _0}

Incorrect

Option 2)

\frac{q}{8\varepsilon _0}

Correct

Option 3)

\frac{q}{\varepsilon _0}

Incorrect

Option 4)

\frac{q}{2\varepsilon _0}

Incorrect

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