QJEEM23Ph151a: JEE Main 2023 Physics - Class 12 JEEM23Ph151a Solution: A capacitor of capacitance is charged to a potential V. The…

The electric field inside a parallel plate capacitor is uniform and given by $\mathbf{E}=\frac{V}{d} \hat{\mathbf{j}}$ where $d$ is the separation between the plates.

The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by $\Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}$ .

Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by $\Phi_E = E A$ , where $A$ is the area of the plate.

The area of the positive plate is $A = \frac{Q}{\varepsilon_0 V}$ , where $Q = C V$ is the charge on the positive plate. Therefore, the electric flux through the surface is given by:

$\Phi_E = \frac{Q}{\varepsilon_0 V} \frac{V}{d} = \frac{Q}{\varepsilon_0 d} = \frac{C V}{\varepsilon_0 d}$

Thus, the answer is $\frac{C V}{\varepsilon_0}$ .