Four charges 2 mu C, -3 mu C, 4 mu C, -4 mu C | vedclass

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

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$\frac{2}{\epsilon_0}$

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(A) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.Here,the enclosed charges are $q_1 = 2 \mu C, q_2 = -3 \mu C, q_3 = 4 \mu C, q_4 = -4 \mu C$,and $q_5 = -1 \mu C$.The net enclosed charge $q_{net} = q_1 + q_2 + q_3 + q_4 + q_5$.$q_{net} = (2 - 3 + 4 - 4 - 1) \mu C = -2 \mu C$.Therefore,the net outward flux $\phi = \frac{-2 \mu C}{\epsilon_0} = -\frac{2}{\epsilon_0} \mu V-m$.Note: Since the question asks for the magnitude or the value based on the provided options,and the sum is $-2 \mu C$,there might be a typo in the question's provided options. However,based on the calculation,the net flux is $-\frac{2}{\epsilon_0}$.

Four charges $2 \mu C, -3 \mu C, 4 \mu C, -4 \mu C$ and $-1 \mu C$ are enclosed by a Gaussian surface of radius $2 \ m$. The net outward flux through the Gaussian surface is (in $\mu V-m$):

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