Sea water at frequency $\nu = 4 \times 10^8\ \text{Hz}$ has permittivity $\varepsilon = 80\,\varepsilon_0$, permeability $\mu = \mu_0$ and resistivity $\rho = 0.25\ \Omega\text{m}$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t) = V_0\sin(2\pi\nu t)$. What fraction of the conduction current density is the displacement current density?

Sea water at frequency nu = 4 × 10⁸Hz has permittivity varepsilon = 80varepsilon₀, permeability μ = μ₀ and res

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\nu = 4 \times 10^8\ \text{Hz}

Assam AhsecSubjective· 3mImportance★★★★★

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Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Ampere's circuital law, in its original form, links the magnetic field around a closed loop to the conduction current (moving charges) threading that loop:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c

Maxwell realised this law is incomplete. The classic illustration is a charging capacitor. Consider an Amperian loop encircling the wire that feeds one plate.

If you cap that loop with a flat surface cut by the wire, a real conduction current $I_c$ passes through it.
If you instead cap the SAME loop with a bulging surface that passes between the two capacitor plates, no charge crosses the gap — the space between the plates is an insulator. So $I_c = 0$ through this surface.

Ampere's law now gives two different answers for $\oint \vec{B}\cdot d\vec{l}$ for the same loop, depending on which surface you choose. That is a contradiction — the law cannot be right as it stands.

Maxwell's Fix: A Current Made of Changing Field

Between the plates there is no moving charge, but there is a growing electric field, because charge is piling up on the plates. Maxwell proposed that a changing electric flux acts like a current for the purpose of producing a magnetic field. He called it the displacement current, $I_d$ .

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

where $\Phi_E = \int \vec{E}\cdot d\vec{A}$ is the electric flux through the surface, and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space.

Check with the capacitor. For a parallel-plate capacitor of area $A$ and plate charge $q$ , the field between the plates is $E = \dfrac{q}{\varepsilon_0 A}$ , so the flux is $\Phi_E = EA = \dfrac{q}{\varepsilon_0}$ . Then

I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{1}{\varepsilon_0}\frac{dq}{dt} = \frac{dq}{dt} = I_c

So the displacement current in the gap is exactly equal to the conduction current in the wire. The two surfaces now give the same answer — the contradiction is gone.

The Complete (Ampere–Maxwell) Law

Maxwell rewrote Ampere's law so that the total current is conduction plus displacement current:

\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I_c + I_d\right) = \mu_0 I_c + \mu_0\varepsilon_0 \frac{d\Phi_E}{dt}

Important

The deep meaning: a changing electric field produces a magnetic field, just as (by Faraday's law) a changing magnetic field produces an electric field. This symmetry is what makes self-sustaining electromagnetic waves possible — the changing E-field of the wave generates the B-field and vice versa.

Key Points to Remember …

Why this formula?

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:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

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.

Surface	Current through it
S₁ (cuts wire)	$I$
S₂ (between plates)	$0$

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:

\Phi_E = \int \vec{E} \cdot d\vec{A} = E \cdot A = \frac{Q}{\varepsilon_0}

Step 2: Differentiate with respect to time

\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dQ}{dt} = \frac{I}{\varepsilon_0}

Step 3: Define displacement current

Maxwell defined the displacement current $I_d$ as:

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

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:

\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_{\text{enc}} + I_d)

Or equivalently:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}

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

Quantity	Formula	Meaning
Displacement current	$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$	Equivalent "current" from changing E-field

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The required fraction is $\dfrac{J_d}{J_c}=\dfrac{\varepsilon\omega}{\sigma}\approx0.445$ : at this frequency the displacement current density is about $44.5\%$ of the conduction current density.

Setting up the two current densities. Inside the capacitor the same electric field $E(t)$ drives both a conduction current (moving ions in the sea water) and a displacement current (the changing field in the medium):

J_c=\sigma E,\qquad J_d=\varepsilon\frac{\partial E}{\partial t}.

Because both are produced by the same field, their ratio does not depend on the plate area, the separation, or $V_0$ — only on the material properties and the frequency.

Step 1 — Conductivity from resistivity.

\sigma=\frac{1}{\rho}=\frac{1}{0.25\ \Omega\text{m}}=4\ \text{S/m}.

Step 2 — Time dependence. The source gives $V(t)=V_0\sin(2\pi\nu t)$ , so the field is $E(t)=E_0\sin(\omega t)$ with $\omega=2\pi\nu$ . Then

\frac{\partial E}{\partial t}=\omega E_0\cos(\omega t),

so the peak current densities are $J_c^{\max}=\sigma E_0$ and $J_d^{\max}=\varepsilon\omega E_0$ .

Step 3 — Form the ratio.

\frac{J_d}{J_c}=\frac{\varepsilon\omega E_0}{\sigma E_0}=\frac{\varepsilon\omega}{\sigma}=\frac{80\varepsilon_0\,(2\pi\nu)}{\sigma}.

Step 4 — Put in the numbers.

\varepsilon=80\times8.85\times10^{-12}=7.08\times10^{-10}\ \text{F/m},

$\omega=2\pi\times4\times10^{8}=2.51\times10^{9}\ \text{rad/s}.$ …

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Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Maxwell's Fix: A Current Made of Changing Field

The Complete (Ampere–Maxwell) Law

Key Points to Remember …

Why this formula?

Displacement Current: Why the Formula Holds

1. The Problem That Demanded a Fix

2. The Insight: Changing Electric Field

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

Step 2: Differentiate with respect to time

Step 3: Define displacement current

4. The Corrected Ampère-Maxwell Law

5. The Key Formula(e) — Summarized

Method: Comparing Conduction and Displacement Current Density in a Lossy Dielectric

Steps

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