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

Answer 1

The correct option is D

${\epsilon }_o({\varphi }_2-{\varphi }_1)$

Step1: Given data

Electric flux ${\Phi }_1$ (or electric field lines) entering a closed surface is negative, whereas electric flux ${\Phi }_2$ (or electric field lines) exiting a closed surface is positive.
As a result, net electric flux, $\Phi =({\Phi }_2-{\Phi }_1)$

Step2: Gauss's law

1. The total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity, according to Gauss Law.
2. The total flux associated with a closed surface equals $\frac{1}{{\epsilon }_o}$ times the charge encompassed by the closed surface, according to the Gauss law.

Step3: Applying gauss law

$$\begin{gathered} \Phi =\frac{q_{in}}{{\epsilon }_o} \\ {q}_{in}={\epsilon }_o\Phi \\ {q}_{in}=({\Phi }_2-{\Phi }_1){\epsilon }_o \end{gathered}$$

Hence, option D is the correct answer.