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

Answers (1)

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}}

Posted by

Satyajeet Kumar

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