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Question

(B) According to Gauss's Law,the net electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free

Vedclass · Accepted Answer

$(\varphi _2 - \varphi _1)\varepsilon _0$

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(B) According to Gauss's Law,the net electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space: $\Phi_{net} = \frac{Q_{enc}}{\varepsilon_0}$.Here,the flux entering the surface is $\varphi_1$ (negative flux) and the flux leaving the surface is $\varphi_2$ (positive flux).Therefore,the net flux is $\Phi_{net} = \varphi_2 - \varphi_1$.Substituting this into Gauss's Law: $\varphi_2 - \varphi_1 = \frac{Q_{enc}}{\varepsilon_0}$.Thus,the charge enclosed is $Q_{enc} = (\varphi_2 - \varphi_1)\varepsilon_0$.

If the electric flux entering and leaving an enclosed surface respectively is ${\varphi _1}$ and ${\varphi _2}$,the electric charge inside the surface will be:

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