Four closed surfaces and corresponding charge | vedclass

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(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$,where $q_{e

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$\phi_1 = \phi_2 = \phi_3 = \phi_4$

Vedclass · Answer

(C) According to Gauss's Law,the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$,where $q_{enclosed}$ is the total charge enclosed by the surface.For surface $S_1$: The enclosed charge is $2q$. Therefore,$\phi_1 = \frac{2q}{\varepsilon_0}$.For surface $S_2$: The enclosed charges are $q, q, q, -q$. The net enclosed charge is $q + q + q - q = 2q$. Therefore,$\phi_2 = \frac{2q}{\varepsilon_0}$.For surface $S_3$: The enclosed charges are $q, q$. The charge $5q$ is outside the surface,so it does not contribute to the flux. The net enclosed charge is $q + q = 2q$. Therefore,$\phi_3 = \frac{2q}{\varepsilon_0}$.For surface $S_4$: The enclosed charges are $8q, -2q, -4q$. The charge $3q$ is outside the surface. The net enclosed charge is $8q - 2q - 4q = 2q$. Therefore,$\phi_4 = \frac{2q}{\varepsilon_0}$.Comparing the results,we find that $\phi_1 = \phi_2 = \phi_3 = \phi_4 = \frac{2q}{\varepsilon_0}$.Thus,the net electric flux is the same for all surfaces.

Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be $\phi_1, \phi_2, \phi_3$ and $\phi_4$. Then:

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