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 :
- \(\frac{CV}{2\epsilon_0}\)
- \(\frac{2CV}{\epsilon_0}\)
- Zero
- \(\frac{CV}{\epsilon_0}\)
The Correct Option is A
Solution and Explanation
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:
\[ \Phi = \frac{C \cdot V}{\varepsilon_0}. \]
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