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_2 + \phi_1 }{ \varepsilon_0}$
• C. $ \frac{ \phi_1 - \phi_2 }{ \varepsilon_0}$
• D. $ \varepsilon_0 ( \phi_1 + \phi_2 ) $

Answer 1

Answer: (D)

Net electric flux,$\phi=\phi_{2}-\phi_{1}$
Applying Gauss's law, $\phi=\int \overrightarrow{\text { E.fnds }}=\frac{Q_{\text {enclosed }}}{\epsilon_{0}}$
So, $Q _{\text {enclosed }}=\phi \epsilon_{0}=\left(\phi_{2}-\phi_{1}\right) \epsilon_{0}$