A capacitor of capacitance `C` is charged to a potential `V`. The flux of the electric field through a closed surface enclosing the capacitor is

Options

A. `(CV)/(epsilon)`
B. `(2 CV)/(epsilon_0)`
C. `(CV)/(2espsilon_0)`
D. zero

Answer

To solve the problem, we need to determine the electric flux through a closed surface that encloses a charged capacitor. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Electric Flux
Electric flux (Φ) through a closed surface is defined by Gauss's law, which states that the electric flux through a closed surface is equal to the total charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀):
\[ \Phi = \frac{Q}{\epsilon_0} \]

### Step 2: Identify the Charges in the Capacitor
A capacitor consists of two plates: one with a positive charge (+Q) and the other with an equal but negative charge (-Q). When the capacitor is charged to a potential V, the total charge on the capacitor is Q.

### Step 3: Determine the Total Charge Enclosed
When we consider a closed surface that completely encloses the capacitor, we need to calculate the total charge enclosed by that surface. Since the positive charge on one plate is +Q and the negative charge on the other plate is -Q, the total charge enclosed (Q_enclosed) is:
\[ Q_{\text{enclosed}} = +Q + (-Q) = 0 \]

### Step 4: Apply Gauss's Law
Now we can substitute the total charge enclosed into the equation for electric flux:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} = \frac{0}{\epsilon_0} = 0 \]

### Conclusion
Thus, the electric flux through a closed surface enclosing the capacitor is:
\[ \Phi = 0 \]

### Final Answer
The correct option is **0**.

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