Question 220: Electrostatics & Capacitance

Question

A capacitor of capacitance $$\mathrm{C}$$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :

Accepted Answer

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 $\frac{C V}{\varepsilon_0}$.

Answer

Zero

Answer

$$\frac{C V}{\varepsilon_{0}}$$

Answer

$$\frac{C V}{2 \varepsilon_{0}}$$

Answer

$$\frac{2 C V}{\varepsilon_{0}}$$

Electrostatics & Capacitance

A capacitor of capacitance $\mathrm{C}$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :

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