Question

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

Answer 1

Answer: (D)

According to this law, the net electric flux through any dosed surface is equal to the net charge inside the surface divided by ${\epsilon }_0$''.
Therefore, $\varphi =\frac{q}{{\epsilon }_0}$
Let $-q_1$ be the charge, due to which flux ${\varphi }_1$ is entering the surface
$\therefore \varphi =\frac{-q_1}{{\epsilon }_0}$
or $-q_1={\epsilon }_0{\varphi }_1$
Let $+q_2$ be the charge, due to which flux ${\varphi }_2$ is leaving, the surface
$\therefore {\varphi }_2=\frac{q_2}{{\epsilon }_0}$
or $q_2={\epsilon }_0{\varphi }_2$
So, electric charge inside the surface
$=q_2-q_1$
$={\epsilon }_0{\varphi }_2+{\epsilon }_0{\varphi }_1$
$={\epsilon }_0\left({\varphi }_2+{\varphi }_1\right)$