If there is only one type of charge in the un | vedclass

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(D) According to Gauss's Law,the electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivi

Vedclass · Accepted Answer

$ \oint \vec{E} \cdot d\vec{s} = 0 $ if charge is outside,$ = \frac{q}{\varepsilon_{0}} $ if charge is inside

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(D) According to Gauss's Law,the electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space $( \varepsilon_{0} )$.Mathematically,$ \oint \vec{E} \cdot d\vec{s} = \frac{q_{	ext{enclosed}}}{\varepsilon_{0}} $.If the charge is outside the Gaussian surface,the net charge enclosed $( q_{	ext{enclosed}} )$ is $ 0 $,so the flux $ \oint \vec{E} \cdot d\vec{s} = 0 $.If the charge is inside the Gaussian surface,the net charge enclosed is $ q $,so the flux $ \oint \vec{E} \cdot d\vec{s} = \frac{q}{\varepsilon_{0}} $.This law holds true regardless of whether there is only one type of charge or both types of charges in the universe.

If there is only one type of charge in the universe,then ($ \vec{E} $ = Electric field,$ \vec{d}s $ = Area vector):

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