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

Question

If there are only one type of charge in the universe, then ( $ \vec{E} $ Electric field, $ \vec{d} s \rightarrow $ Area vector $ ) $, $ \oint \vec{E} \cdot \overrightarrow{d_{S}} \neq 0 $ on any surface, $ \oint \vec{E} \cdot \vec{d} s $ could not be defined, $ \oint \vec{E} \cdot d s=\infty $ if charge is inside, $ \oint \vec{E} \cdot d s=0 $ if charge is outside, $ =\frac{q}{\varepsilon_{0}} $ if charge is inside

MARKS App · Accepted Answer

If there are only one type of charge in the universe, then by Gauss Law: if charge is outside the surface then charge enclosed by gaussian surface is zero. Therefore, $ \oint \vec{E} \cdot d \vec{s}=0 $ If charge is inside the gaussian surface, then flux $ \phi $ is given by $ \oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}} $

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