When a 10 mu C charge is enclosed by a closed | vedclass

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(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.Initially,the c

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$2 \phi$

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(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.Initially,the charge enclosed is $q = 10 \mu C$,so $\phi = \frac{q}{\varepsilon_0} \quad ... (1)$.When another $10 \mu C$ charge is placed inside,the total charge enclosed becomes $Q = 10 \mu C + 10 \mu C = 20 \mu C = 2q$.The new flux $\phi'$ passing through the surface is $\phi' = \frac{Q}{\varepsilon_0} = \frac{2q}{\varepsilon_0}$.Substituting from equation $(1)$,we get $\phi' = 2 \phi$.

When a $10 \mu C$ charge is enclosed by a closed surface,the flux passing through the surface is $\phi$. Now,another $10 \mu C$ charge is placed inside the closed surface,then the flux passing through the surface is . . . . . . .

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