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Question

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

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$10^3\, N m^2 C^{-1}$

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(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.Here,the Gaussian surface $A$ encloses all three charges $q_1$,$q_2$,and $q_3$.Total enclosed charge $q_{enclosed} = q_1 + q_2 + q_3 = (-14 + 78.85 - 56)\, nC = 8.85\, nC = 8.85 	imes 10^{-9}\, C$.The value of permittivity of free space is $\varepsilon_0 \approx 8.85 	imes 10^{-12}\, C^2 N^{-1} m^{-2}$.Therefore,the electric flux $\phi = \frac{8.85 	imes 10^{-9}}{8.85 	imes 10^{-12}} = 10^3\, N m^2 C^{-1}$.

The electric flux for Gaussian surface $A$ that encloses the charged particles in free space is (given $q_1 = -14\, nC$,$q_2 = 78.85\, nC$,$q_3 = -56\, nC$):

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