If the electric flux entering and leaving a c | vedclass

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Question

(B) According to the sign convention for electric flux,flux entering a closed surface is considered negative $(- \phi_1)$ and flux leaving a closed su

Vedclass · Accepted Answer

$(\phi_2 - \phi_1)\varepsilon_0$

Vedclass · Answer

(B) According to the sign convention for electric flux,flux entering a closed surface is considered negative $(- \phi_1)$ and flux leaving a closed surface is considered positive $(+ \phi_2)$.The net electric flux $\phi_{	ext{net}}$ through the closed surface is given by the sum of the fluxes:$\phi_{	ext{net}} = \phi_2 - \phi_1$According to Gauss's Law,the net electric flux through a closed surface is equal to $\frac{1}{\varepsilon_0}$ times the total charge $q_{	ext{in}}$ enclosed by the surface:$\phi_{	ext{net}} = \frac{q_{	ext{in}}}{\varepsilon_0}$Substituting the value of $\phi_{	ext{net}}$:$\phi_2 - \phi_1 = \frac{q_{	ext{in}}}{\varepsilon_0}$Therefore,the total charge inside the surface is:$q_{	ext{in}} = (\phi_2 - \phi_1)\varepsilon_0$

If the electric flux entering and leaving a closed surface are $\phi_1$ and $\phi_2$ respectively,then the electric charge inside the surface will be:

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