Question

If the electric flux entering and leaving an enclosed surface respectively are $\phi_{1}$ and $\phi_{2}$, the electric charge inside the surface will be

• A. $\frac{\phi_{2}-\phi_{1}}{\varepsilon_{0}}$
• B. $\frac{\phi_{1}+\phi_{2}}{\varepsilon_{0}}$
• C. $\frac{\phi_{1}-\phi_{2}}{\varepsilon_{0}}$
• D. $\varepsilon_{0}\left(\phi_{1}+\phi_{2}\right)$

Answer 1

Answer: (D)

$\varepsilon_{0}\left(\phi_{1}+\phi_{2}\right)$

Answer 2

According to this law, the net electric flux through any dosed surface is equal to the net charge inside the surface divided by $\varepsilon_{0}$''.
Therefore, $\phi=\frac{q}{\varepsilon_{0}}$
Let $-q_{1}$ be the charge, due to which flux $\phi_{1}$ is entering the surface
$\therefore \phi=\frac{-q_{1}}{\varepsilon_{0}}$
or $-q_{1}=\varepsilon_{0} \phi_{1}$
Let $+q_{2}$ be the charge, due to which flux $\phi_{2}$ is leaving, the surface
$\therefore \phi_{2}=\frac{q_{2}}{\varepsilon_{0}}$
or $q_{2}=\varepsilon_{0} \phi_{2}$
So, electric charge inside the surface
$=q_{2}-q_{1}$
$=\varepsilon_{0} \,\phi_{2}+\varepsilon_{0} \,\phi_{1}$
$=\varepsilon_{0}\left(\phi_{2}+\phi_{1}\right)$