Question

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. \(\frac{CV}{2\epsilon_0}\)
• B. \(\frac{2CV}{\epsilon_0}\)
• C. Zero
• D. \(\frac{CV}{\epsilon_0}\)

Answer 1

The Correct Option is A

Step 1: Recall Gauss’s Law

Gauss’s law states that the electric flux (\( \Phi \)) through a closed surface is proportional to the enclosed charge (\( q \)) and inversely proportional to the permittivity of free space (\( \varepsilon_0 \)):

\[ \Phi = \frac{q}{\varepsilon_0}. \]

Step 2: Relate Charge, Capacitance, and Potential Difference

The charge (\( q \)) on a capacitor is related to its capacitance (\( C \)) and the potential difference (\( V \)) across it by the formula:

\[ q = C \cdot V. \]

Step 3: Calculate the Electric Flux

Substitute \( q = C \cdot V \) into Gauss’s law:

\[ \Phi = \frac{q}{\varepsilon_0} = \frac{C \cdot V}{\varepsilon_0}. \]

Final Answer:

The flux of the electric field through the closed surface is:

\[ \Phi = \frac{C \cdot V}{\varepsilon_0}. \]