What is the electric flux for Gaussian surfac | vedclass

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Question

(A) According to Gauss's law of electrostatics,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varep

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) According to Gauss's law of electrostatics,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.The net charge $q$ enclosed by the Gaussian surface $A$ is the sum of the individual charges:$q = q_1 + q_2 + q_3$$q = (-14 + 78.85 - 56) 	ext{ nC} = 8.85 	ext{ nC} = 8.85 	imes 10^{-9} 	ext{ C}$.The permittivity of free space is $\varepsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$.Substituting these values into the formula for electric flux:$\phi = \frac{8.85 	imes 10^{-9} 	ext{ C}}{8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}}$$\phi = 10^3 	ext{ N m}^2 	ext{ C}^{-1}$.

List-$I$	List-$II$
$(a)$ Net flux through $S_1$	$(i)$ $\frac{-2Q}{\epsilon_0}$
$(b)$ Net flux through $S_2$	(ii) $\frac{-Q}{\epsilon_0}$
$(c)$ Net flux through $S_3$	(iii) Zero

What is the electric flux for Gaussian surface $A$ that encloses the charged particles in free space? [Given: $q_1 = -14 \text{ nC}, q_2 = 78.85 \text{ nC}, q_3 = -56 \text{ nC}$]

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