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

Answer 1

Answer: (D)

Net electric flux,$\varphi ={\varphi }_2-{\varphi }_1$
Applying Gauss's law, $\varphi =\int \overrightarrow{\text{ E.fnds }}=\frac{Q_{\text{enclosed }}}{{ϵ}_0}$
So, $Q_{\text{enclosed }}=\varphi {ϵ}_0=\left({\varphi }_2-{\varphi }_1\right){ϵ}_0$