Three charges q_1 = 1,mu C, q_2 = 2,mu C,and | vedclass

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

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$-36\pi 	imes 10^3$

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Here,$q_{	ext{enclosed}}$ is the net charge enclosed by the surface $S_2$.From the figure,the surface $S_2$ encloses charges $q_2$ and $q_3$.Therefore,$q_{	ext{enclosed}} = q_2 + q_3 = 2\,\mu C + (-3\,\mu C) = -1\,\mu C = -1 	imes 10^{-6}\,C$.Using the relation $\frac{1}{\epsilon_0} = 4\pi k$,where $k = 9 	imes 10^9\,N\cdot m^2/C^2$:$\phi = 4\pi k 	imes q_{	ext{enclosed}}$$\phi = 4\pi 	imes (9 	imes 10^9) 	imes (-1 	imes 10^{-6})$$\phi = -36\pi 	imes 10^3\,N\cdot m^2/C$.

Three charges $q_1 = 1\,\mu C, q_2 = 2\,\mu C$,and $q_3 = -3\,\mu C$ and four surfaces $S_1, S_2, S_3$,and $S_4$ are shown in the figure. The electric flux emerging through surface $S_2$ in $N\cdot m^2/C$ is:

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