Q.A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius $R = 6.0\ \text{cm}$ has a capacitance $C = 100\ \text{pF}$. The capacitor is connected to a $230\ \text{V}$ ac supply with a (angular) frequency of $300\ \text{rad s}^{-1}$.

Displacement Current: Why the Formula Holds

The displacement current is one of the most elegant corrections in physics — it fixed a logical flaw in Maxwell's equations and predicted electromagnetic waves. Let's understand why its formula emerges.

1. The Problem That Demanded a Fix

Consider a capacitor being charged in a circuit. Ampère's law (in its original form) states:

where $I_{\text{enc}}$ is the current passing through any surface bounded by the loop.

Now take two different surfaces bounded by the same loop:

• Surface S₁: Cuts the wire — current $I$ passes through.
• Surface S₂: Passes between the capacitor plates — no current passes through.

This is a contradiction: the same loop gives two different values for $\oint \vec{B} \cdot d\vec{l}$. Ampère's law is inconsistent for time-varying fields.

2. The Insight: Changing Electric Field

Between the capacitor plates, there is no conduction current, but there is a changing electric field as charge builds up.

• The electric field between plates: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$
• As $Q$ changes, $E$ changes: $\frac{dE}{dt} = \frac{1}{\varepsilon_0 A} \frac{dQ}{dt}$

Maxwell realized: a changing electric field should produce a magnetic field, just like a current does.

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

The electric flux through the capacitor plates is:

Step 2: Differentiate with respect to time

Step 3: Define displacement current

Maxwell defined the displacement current $I_d$ as:

From Step 2, this equals $I$ — the same conduction current in the wire. The displacement current "bridges" the gap.

4. The Corrected Ampère-Maxwell Law

The full law becomes:

Or equivalently:

Why this works:

• For surface S₁: $I_{\text{enc}} = I$, $\frac{d\Phi_E}{dt} = 0$ → result = $\mu_0 I$
• For surface S₂: $I_{\text{enc}} = 0$, $\frac{d\Phi_E}{dt} = \frac{I}{\varepsilon_0}$ → result = $\mu_0 \varepsilon_0 \cdot \frac{I}{\varepsilon_0} = \mu_0 I$

Both surfaces give the same answer. The contradiction is resolved.

5. The Key Formula(e) — Summarized

6. Why This Matters for Exams

• Conceptual trap: Students often think displacement current is a real current of charges. It is not — it's a term that behaves like a current in producing magnetic fields.
• Numerical problems: You'll often compute $I_d$ from $\frac{dE}{dt}$ or from the charging rate of a capacitor.
• Key exam point: The displacement current is zero in steady-state DC circuits (constant fields), but non-zero in AC circuits or during charging/discharging.

7. The Deeper "Why"

The displacement current isn't just a mathematical patch — it reveals a profound symmetry:

• A changing magnetic field produces an electric field (Faraday's law)
• A changing electric field produces a magnetic field (Maxwell's correction)

This symmetry is what makes electromagnetic waves possible: each changing field sustains the other, allowing energy to propagate through empty space.

Final takeaway: The formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ holds because it makes Ampère's law consistent for all surfaces and reveals the deep symmetry between electricity and magnetism.