Question

An AC generator with output voltage \(V\) and frequency \(f\) is connected to the plates of an air-filled parallel plate capacitor of plate area \(A\) and plate separation \(d\). The maximum value of the displacement current is:

• A. \(\dfrac{2\pi \epsilon_0 f V A}{d}\)
• B. \(\dfrac{\pi \epsilon_0 f V A}{d}\)
• C. \(\dfrac{2\pi \epsilon_0 f A}{V d}\)
• D. \(\dfrac{2\pi \epsilon_0 V A}{f d}\)

Hint:

Use \(I_d = C\,dV/dt\) with \(C = \epsilon_0 A/d\) and peak of \(dV/dt = 2\pi f V\).

Answer 1

The Correct Option is A

Step 1: View the capacitor as a circuit element of capacitance $C = \dfrac{\epsilon_0 A}{d}$. The displacement current inside the gap is identical to the charging current $I = \dfrac{dQ}{dt}$ delivered by the generator.

Step 2: The charge on the capacitor is $Q = C V(t)$. With $V(t) = V\sin(2\pi f t)$,
\[Q = C V \sin(2\pi f t)\]
Step 3: Differentiate to get the current:
\[I = \frac{dQ}{dt} = C V (2\pi f)\cos(2\pi f t)\]
Step 4: The peak (maximum) current is the amplitude of this cosine:
\[I_{\max} = 2\pi f C V = 2\pi f V \cdot \frac{\epsilon_0 A}{d}\]
Step 5: Rearranging gives
\[I_{\max} = \frac{2\pi \epsilon_0 f V A}{d}\]
identical to the field-flux derivation, so option (A) is correct.
\[\boxed{I_{\max} = \dfrac{2\pi \epsilon_0 f V A}{d}}\]