If the electric flux entering and leaving an | vedclass

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(A) According to Gauss's Law, the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface div

Vedclass · Accepted Answer

$(\phi_2 - \phi_1)\varepsilon_0$

Vedclass · Answer

(A) According to Gauss's Law, the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.The net flux is given by the difference between the flux leaving the surface and the flux entering the surface:$\phi_{net} = \phi_{out} - \phi_{in} = \phi_2 - \phi_1$Applying Gauss's Law:$\phi_{net} = \frac{q}{\varepsilon_0}$Substituting the net flux:$\phi_2 - \phi_1 = \frac{q}{\varepsilon_0}$Therefore, the charge $q$ inside the surface is:$q = \varepsilon_0(\phi_2 - \phi_1)$

If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be:

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