MHT CET · Physics · Electrostatics
If the electric flux entering and leaving an enclosed surface is \(\phi_1\) and \(\phi_2\) then charge enclosed in the surface is ( \(\varepsilon_0=\) permittivity of free space)
- 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 \(\varepsilon_0\left(\phi_2-\phi_1\right)\)
Answer & Solution
Correct Answer
(D) \(\varepsilon_0\left(\phi_2-\phi_1\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \Phi=\frac{q_{i n}}{\varepsilon_o} \\
& q_{i n}=\varepsilon_o \Phi \\
& q_{\text {in }}=\left(\Phi_2-\Phi_1\right) \varepsilon_o
\end{aligned}\)
Hence, option D is the correct answer.
& \Phi=\frac{q_{i n}}{\varepsilon_o} \\
& q_{i n}=\varepsilon_o \Phi \\
& q_{\text {in }}=\left(\Phi_2-\Phi_1\right) \varepsilon_o
\end{aligned}\)
Hence, option D is the correct answer.
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