Q98 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
A capacitor of capacitance $C$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is:
- A$\frac{C V}{2 \varepsilon_0}$
- B$\frac{2 \mathrm{CV}}{\varepsilon_0}$
- Correct answerC$\frac{\mathrm{CV}}{\varepsilon_0}$
- D
Zero
Solution
From the Gauss's law
$\begin{aligned}<br />& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{Q}}{\epsilon_0} \quad \because \mathrm{Q}=\mathrm{CV} \\<br />& \therefore \phi=\frac{\mathrm{CV}}{\epsilon_0}<br />\end{aligned}$
$\begin{aligned}<br />& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{Q}}{\epsilon_0} \quad \because \mathrm{Q}=\mathrm{CV} \\<br />& \therefore \phi=\frac{\mathrm{CV}}{\epsilon_0}<br />\end{aligned}$
Hence, the correct answer is option: 'C'.