MHT CET · Physics · Electrostatics
The electric flux linked with the closed surface in \(\mathrm{Nm}^2 \mathrm{C}^{-1}\) is
\(\left(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\)

- A \(10^{12}\)
- B \(8.85 \times 10^{-13}\)
- C \(10^{10}\)
- D \(10^{11}\)
Answer & Solution
Correct Answer
(A) \(10^{12}\)
Step-by-step Solution
Detailed explanation
Electric flux, \(\phi=\frac{q}{\varepsilon_0}\)
where \(q=\) total charge enclosed by closed surface
\(q=(2.35+5+2-0.5) C\)
\(q=(2.35+5+2-0.5) \mathrm{C}\)
\(\phi=\frac{q}{\varepsilon_0}=\frac{8.85 \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=10^{12} \mathrm{Nm}^2 / \mathrm{C}\)
where \(q=\) total charge enclosed by closed surface
\(q=(2.35+5+2-0.5) C\)
\(q=(2.35+5+2-0.5) \mathrm{C}\)
\(\phi=\frac{q}{\varepsilon_0}=\frac{8.85 \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}}=10^{12} \mathrm{Nm}^2 / \mathrm{C}\)
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