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)

$\varepsilon_0 ( \phi_1 + \phi_2 )$

Answer 2

According to Gauss theorem, "the net electric flux through any closed surface is equal to the net charge inside the surface divided by $\varepsilon_0$'' . Therefore, $\phi = \frac{ q}{ \varepsilon_0 }$ Let -$q_1$ be the charge, due to which flux $\phi$ is entering the surface $\phi_1 = \frac{- q_1}{ \varepsilon_0 }$ or $-q_1 = \varepsilon_0 \phi_1$ Let $+ q_2$ be the charge, due to which flux $\phi_2$ is leaving the surface $\phi_2 = \frac{ q_2}{ \varepsilon_0 }$ or $-q_2 = \varepsilon_0 \phi_2$ So, electric charge inside the surface = $q_2 - q_1$ = $\varepsilon_0 \phi_2 + \varepsilon_0 \phi_1 = \varepsilon_0 (\phi_2 + \phi_1 )$