Question

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

• A. ${\epsilon }_0({\varphi }_1+{\varphi }_2)$
• B. $\frac{{\varphi }_2-{\varphi }_1}{{\epsilon }_0}$
• C. $\frac{{\varphi }_1-{\varphi }_2}{{\epsilon }_0}$
• D. $\frac{{\varphi }_1+{\varphi }_2}{{\epsilon }_0}$

Answer 1

The correct option is A ${\epsilon }_0({\varphi }_1+{\varphi }_2)$

According to Gauss theorem, "the net electric flux through any closed surface is equal to the net charge inside the surface divided by ${\epsilon }_0$".
Therefore, $\varphi =\frac{q}{{\epsilon }_0}$
Let $-q$ be the charge, due to which the flux $\varphi$, is entering the surface, ${\varphi }_1=-\frac{q_1}{{\epsilon }_0}$
$\Rightarrow -q_1={\epsilon }_0{\varphi }_1$
Let $q_2$ be the charge, due to which the flux ${\varphi }_2$ is leaving the surface.
$\therefore {\varphi }_2=\frac{q_2}{{\epsilon }_0}\Rightarrow q={\epsilon }_0{\varphi }_2$
So, charge inside the surface
$=q_2-q_1$
$={\epsilon }_0{\varphi }_2+{\epsilon }_0{\varphi }_1={\epsilon }_0({\varphi }_1+{\varphi }_2)$