Q173 / 225 · JEE Main 2003 · Physics

If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be

  • Correct answer
    A
    $\left(\phi_2-\phi_1\right) \varepsilon_0$
  • B
    $\left(\phi_1+\phi_2\right) / \varepsilon_0$
  • C
    $\left(\phi_2-\phi_1\right) / \varepsilon_0$
  • D
    $\left(\phi_1+\phi_2\right) \varepsilon_0$

Solution

According to Gauss's Law $$ \begin{aligned} & \int(\text { E.dA })=\mathrm{q}_0 / \varepsilon_0 \Rightarrow \mathrm{q}=\varepsilon_0\left(\phi_2-\phi_1\right) \\ & {\left[\text { since } \phi=\int \mathrm{E} \cdot \mathrm{dA}\right]} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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