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Question

(D) According to Gauss's Law, the net electric flux $\Phi_{net}$ through a closed surface is equal to the total enclosed charge $q_{in}$ divided by th

Vedclass · Accepted Answer

$\varepsilon_0(\phi_2-\phi_1)$

Vedclass · Answer

(D) According to Gauss's Law, the net electric flux $\Phi_{net}$ through a closed surface is equal to the total enclosed charge $q_{in}$ divided by the permittivity of free space $\varepsilon_0$.$\Phi_{net} = \frac{q_{in}}{\varepsilon_0}$Here, the flux entering the surface is $\phi_1$ (taken as negative) and the flux leaving the surface is $\phi_2$ (taken as positive).Therefore, the net flux is $\Phi_{net} = \phi_2 - \phi_1$.Substituting this into Gauss's Law:$\phi_2 - \phi_1 = \frac{q_{in}}{\varepsilon_0}$$q_{in} = \varepsilon_0(\phi_2 - \phi_1)$.Hence, option $D$ is the correct answer.

If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ respectively, then the charge enclosed in the surface is ($\varepsilon_0 =$ permittivity of free space).

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