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(B) According to Gauss's Law,the electric flux $\phi$ through any closed surface is equal to the total charge enclosed by the surface divided by the p

Vedclass · Accepted Answer

$2q/{\varepsilon _0}$

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(B) According to Gauss's Law,the electric flux $\phi$ through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.Mathematically,$\phi = \frac{Q_{enc}}{\varepsilon_0}$.In the given figure,the surface $S$ encloses two charges,each of magnitude $+q$.Therefore,the total enclosed charge $Q_{enc} = +q + q = 2q$.Substituting this into Gauss's Law,we get $\phi = \frac{2q}{\varepsilon_0}$.

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface $S$ is

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