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Question

(D) According to Gauss's law, the net electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{Q_{enclosed}}{\varepsilon_0}

Vedclass · Accepted Answer

$-4 	imes 10^3 \varepsilon_0 	ext{ C}$

Vedclass · Answer

(D) According to Gauss's law, the net electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{Q_{enclosed}}{\varepsilon_0}$.Inward flux is taken as negative and outward flux is taken as positive.Given: Inward flux $\phi_{in} = -8 	imes 10^3 	ext{ N} \cdot m^2/C$ and outward flux $\phi_{out} = 4 	imes 10^3 	ext{ N} \cdot m^2/C$.The net flux $\phi_{net} = \phi_{in} + \phi_{out} = -8 	imes 10^3 + 4 	imes 10^3 = -4 	imes 10^3 	ext{ N} \cdot m^2/C$.Using Gauss's law: $Q_{enclosed} = \phi_{net} 	imes \varepsilon_0$.Therefore, $Q_{enclosed} = (-4 	imes 10^3) \varepsilon_0 	ext{ C}$.

The inward and outward electric flux for a closed surface in units of $N \cdot m^2/C$ are respectively $8 \times 10^3$ and $4 \times 10^3$. Then the total charge inside the surface is [where $\varepsilon_0$ = permittivity constant].

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