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Question #42709

Difficulty: 8/10 | 🟢 Free

The Question

An ideal parallel-plate capacitor consists of two circular plates of radius $R$ separated by a small distance $d$ ( $d \ll R$ ). An alternating voltage $V(t) = V_0 \cos(\omega t)$ is applied across the plates. Neglecting fringing effects and assuming $\omega R \ll c$ (where $c$ is the speed of light in vacuum and $\epsilon_0$ is the vacuum permittivity), what is the time-averaged total electromagnetic energy stored in the cylindrical region between the plates?

Options

A.
$\frac{\epsilon_0 \pi R^2 V_0^2}{4 d}$
B.
$\frac{\epsilon_0 \pi R^2 V_0^2}{4 d} \left( 1 + \frac{\omega^2 R^2}{8 c^2} \right)$
C.
$\frac{\epsilon_0 \pi R^2 V_0^2}{4 d} \left( 1 + \frac{\omega^2 R^2}{4 c^2} \right)$
D.
$\frac{\epsilon_0 \pi R^2 V_0^2}{2 d} \left( 1 + \frac{\omega^2 R^2}{8 c^2} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Self-Energy of Charged Sphere|Energy Distribution in Standing Waves|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the transition from electrostatics to electrodynamics.

A common trap for students is to assume that the energy of a capacitor is purely electric ( $U = \frac{1}{2}CV^2$ ) and simply time-average it to get Option A.

However, in a time-varying field (AC circuit), the changing electric field produces a displacement current, which inherently induces a magnetic field inside the capacitor.

Students must recognize that the total electromagnetic energy density is $u = \frac{1}{2}\epsilon_0 E^2 + \frac{B^2}{2\mu_0}$ and integrate the dynamic magnetic component over the volume to find the correct total stored energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Self-Energy of Charged Sphere
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free