A charge of 20 mu C is placed inside a close | vedclass

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.Initial flux $\phi

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$4 \ \phi$

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.Initial flux $\phi = \frac{20 \ \mu C}{\epsilon_0}$.After adding $80 \ \mu C$,the total charge becomes $q_{total} = 20 \ \mu C + 80 \ \mu C = 100 \ \mu C$.The new flux is $\phi' = \frac{100 \ \mu C}{\epsilon_0}$.The change in flux is $\Delta \phi = \phi' - \phi = \frac{100 \ \mu C}{\epsilon_0} - \frac{20 \ \mu C}{\epsilon_0} = \frac{80 \ \mu C}{\epsilon_0}$.Since $\phi = \frac{20 \ \mu C}{\epsilon_0}$,we can write $\Delta \phi = 4 	imes \left( \frac{20 \ \mu C}{\epsilon_0} ight) = 4 \phi$.

$A$ charge of $20 \ \mu C$ is placed inside a closed surface. If an additional charge of $80 \ \mu C$ is placed inside the surface,the change in flux is:

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