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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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Verified Answer
The correct answer is:
$\left(\phi_2-\phi_1\right) \varepsilon_0$
$\left(\phi_2-\phi_1\right) \varepsilon_0$
According to Gauss's Law
$$
\begin{aligned}
& \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\
& {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]}
\end{aligned}
$$
$$
\begin{aligned}
& \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\
& {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]}
\end{aligned}
$$
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