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

To solve this question, we need to understand the relationship between the electric field and electric flux for a capacitor.

When a capacitor of capacitance \(C\) is charged to a potential difference \(V\), the charge on the capacitor is given by the formula: 

\(Q = CV\)

The electric flux \(\Phi\) through a closed surface is related to the enclosed charge \(Q_{\text{enclosed}}\) and the permittivity of free space \(\epsilon_0\) by Gauss's Law:

\(\Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0}\)

In this problem, we are interested in the flux through a closed surface enclosing only the positive plate of the capacitor. The crucial concept here is that the charge on the positive plate of the capacitor is explicit, which is \(+\frac{Q}{2}\) if only half of the total charge \(Q\) is considered (as half equivalently charges each plate).

Hence, the charge enclosed \(Q_{\text{enclosed}}\) by the closed surface around the positive plate is:

\(+\frac{CV}{2}\)

Therefore, the flux through the closed surface is:

\(\Phi = \frac{+\frac{CV}{2}}{\epsilon_0} = \frac{CV}{2\epsilon_0}\)

This matches the given correct answer \(\frac{CV}{2\epsilon_0}\).

Moreover, let's rule out other options:

• \(\frac{2CV}{\epsilon_0}\): This option wrongly suggests a higher flux as if enclosed charge was more, which isn't the case.
• Zero: This would imply no enclosed charge, also incorrect.
• \(\frac{CV}{\epsilon_0}\): This suggests the entire charge \(Q\) might be enclosed, which doesn't align with the setup.

Thus, the correct answer is \(\frac{CV}{2\epsilon_0}\).