Maxwell’s Displacement Current

Real-Life Hook

How does your smartphone charge wirelessly? The secret lies in Maxwell’s displacement current! When you place your phone on a charging pad, changing electric fields in the gap between the coils create displacement currents, which in turn generate magnetic fields - transferring energy without any physical contact. Similarly, this concept explains how radio waves carry information through space. Maxwell’s brilliant insight - that changing electric fields create magnetic fields just like electric currents do - revolutionized physics and made modern wireless technology possible. Without displacement current, electromagnetic waves couldn’t exist, and you wouldn’t have Wi-Fi, GPS, or radio communication!

The Problem with Ampere’s Law

Original Ampere’s Circuital Law

For steady currents:

Ampere’s Law (Original)

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$$

Problem: This works perfectly for steady currents, but fails for time-varying situations!

The Capacitor Paradox

Consider a charging capacitor connected to a circuit:

Setup:

• Current I flows in wires
• Capacitor plates separated by gap (no conduction current in gap)
• Apply Ampere’s law with different surfaces

Surface 1 (flat disk between wires):

• Encloses current I
• $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ ✓

Surface 2 (bulges through capacitor gap):

• No conduction current passes through (insulator!)
• $\oint \vec{B} \cdot d\vec{l} = 0$ ✗

Same loop, different answers! This is inconsistent!

The Root Cause

The issue: During charging, electric field between plates changes with time:

$$\frac{d\vec{E}}{dt} \neq 0$$

This changing electric field should contribute to the “effective current”!

Maxwell’s Solution: Displacement Current

The Key Insight

Maxwell realized: A changing electric field is equivalent to a current in producing magnetic fields!

Displacement Current Density

$$\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$

Where:

• $\vec{J}_d$ = displacement current density (A/m²)
• $\epsilon_0$ = permittivity of free space = 8.85 × 10⁻¹² F/m
• $\frac{\partial \vec{E}}{\partial t}$ = rate of change of electric field

Displacement Current

Displacement Current

$$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$$

Where $\Phi_E = \int \vec{E} \cdot d\vec{A}$ = electric flux

For uniform field:

$$I_d = \epsilon_0 A \frac{dE}{dt}$$

where A = area of the surface

Modified Ampere-Maxwell Law

Ampere-Maxwell Law

$$\oint \vec{B} \cdot d\vec{l} = \mu_0(I_c + I_d) = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$$

Where:

• $I_c$ = conduction current
• $I_d$ = displacement current

In differential form:

$$\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$

Interactive Demo: Visualize Displacement Current

See how changing electric fields create displacement currents in capacitors and electromagnetic waves.

Displacement Current in a Charging Capacitor

Parallel Plate Capacitor

Consider a capacitor with:

• Plate area = A
• Separation = d
• Charging current = $I_c$
• Charge at time t = q(t)

Electric field between plates:

$$E = \frac{\sigma}{\epsilon_0} = \frac{q}{\epsilon_0 A}$$

Rate of change of electric field:

$$\frac{dE}{dt} = \frac{1}{\epsilon_0 A} \frac{dq}{dt} = \frac{I_c}{\epsilon_0 A}$$

Displacement current:

$$I_d = \epsilon_0 A \frac{dE}{dt} = \epsilon_0 A \times \frac{I_c}{\epsilon_0 A} = I_c$$

Key Result: In a charging capacitor, displacement current equals conduction current!

$$I_d = I_c$$

This resolves the paradox - surface through the gap “sees” displacement current $I_d = I_c$!

Physical Interpretation

What is Displacement Current?

Important: Displacement current is NOT a flow of charges!

What it is:

• An effective current due to changing electric field
• Produces the same magnetic effect as real current
• Exists in vacuum or insulators (where no charges flow)

Analogy: Just as moving charges (current) create magnetic fields, changing electric fields also create magnetic fields.

Comparison Table

Property	Conduction Current ($I_c$)	Displacement Current ($I_d$)
Cause	Flow of charges	Changing electric field
Medium	Conductors	Vacuum, insulators, conductors
Formula	$I_c = \frac{dq}{dt}$	$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$
Magnetic effect	Creates $\vec{B}$	Creates $\vec{B}$ (same as $I_c$)
Joule heating	Yes (I²R loss)	No heating
Real charge flow	Yes	No

Maxwell’s Equations (Complete Set)

With displacement current, Maxwell completed the set of equations governing electromagnetism:

Maxwell’s Equations (Integral Form)

1. Gauss’s Law: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$

2. Gauss’s Law for Magnetism: $\oint \vec{B} \cdot d\vec{A} = 0$

3. Faraday’s Law: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$

4. Ampere-Maxwell Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Key Symmetry:

• Changing magnetic field → Electric field (Faraday)
• Changing electric field → Magnetic field (Maxwell)

This symmetry leads to electromagnetic waves!

Memory Tricks

“CHANGE Creates Current” for Displacement Current

• CHanging electric field
• Acts as
• Non-material
• Generator of
• Effective current

Think: Even though nothing flows, changing E-field acts like a current!

“I_d = ε₀ dΦ/dt” Mnemonic

“Epsilon Delta Phi Dee-Tee”

• Epsilon (ε₀)
• Delta (d/dt)
• Phi (Φ_E)
• Equals Dee (I_d)

Capacitor Rule: “Conduction = Displacement”

In a charging/discharging capacitor:

$$I_d = I_c$$

• Conduction in wires
• Displacement in gap
• Both are equal!

Maxwell’s Modification: “Plus ε₀ dE/dt”

Original Ampere’s Law + Maxwell’s addition

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I \quad \rightarrow \quad \oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0\epsilon_0 \frac{d\Phi_E}{dt}$$

Common Unit Conversion Mistakes

Permittivity of Free Space (ε₀)

• ε₀ = 8.85 × 10⁻¹² F/m (farads per meter)
• Also written as: 8.85 × 10⁻¹² C²/(N·m²)
• In calculations: Often approximated as $\frac{1}{9 \times 10^9 \times 4\pi}$ F/m

Permeability of Free Space (μ₀)

• μ₀ = 4π × 10⁻⁷ T·m/A (tesla·meter/ampere)
• Also: 4π × 10⁻⁷ H/m (henry per meter)

Speed of Light Relation

$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$

• c = 3 × 10⁸ m/s

Electric Field Units

• E in V/m or N/C
• dE/dt in (V/m)/s = V/(m·s)

Displacement Current Units

• $I_d$ in Amperes (A)
• $J_d$ in A/m² (current density)

Common Error: Forgetting ε₀ has very small magnitude (10⁻¹²), so displacement currents in typical circuits are tiny!

Important Formulas Summary

Displacement Current

$$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 A \frac{dE}{dt}$$

Displacement Current Density

$$J_d = \epsilon_0 \frac{dE}{dt}$$

In Charging Capacitor

$$I_d = I_c$$

(displacement current equals conduction current)

Ampere-Maxwell Law

$$\oint \vec{B} \cdot d\vec{l} = \mu_0(I_c + I_d)$$

Electromagnetic Wave Relation

$$c = \frac{1}{\sqrt{\mu_0\epsilon_0}} = 3 \times 10^8 \text{ m/s}$$

3-Level Practice Problems

Level 1: JEE Main Basics

Problem 1: A parallel plate capacitor with circular plates of radius 5 cm is being charged. The electric field between plates increases at a rate of 10¹² V/(m·s). Calculate the displacement current. (ε₀ = 8.85 × 10⁻¹² F/m)

Solution

Given:

• R = 5 cm = 0.05 m
• dE/dt = 10¹² V/(m·s)

Area of plates:

$$A = \pi R^2 = \pi \times (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2$$

Displacement current:

$$I_d = \epsilon_0 A \frac{dE}{dt}$$ $$I_d = 8.85 \times 10^{-12} \times 7.85 \times 10^{-3} \times 10^{12}$$ $$I_d = 8.85 \times 7.85 = 69.5 \text{ A}$$

Answer: $I_d \approx 70$ A

Problem 2: A capacitor of capacitance 10 μF is connected to a 200 V battery. If the capacitor charges completely in 10⁻³ s, find the average displacement current during charging.

Solution

Given:

• C = 10 μF = 10 × 10⁻⁶ F
• V = 200 V
• t = 10⁻³ s

Charge on capacitor:

$$Q = CV = 10 \times 10^{-6} \times 200 = 2 \times 10^{-3} \text{ C}$$

Average charging current (= average displacement current):

$$I_d = I_c = \frac{Q}{t} = \frac{2 \times 10^{-3}}{10^{-3}} = 2 \text{ A}$$

Answer: $I_d = 2$ A

Note: In a capacitor, $I_d = I_c$ always!

Level 2: JEE Main Advanced

Problem 3: A parallel plate capacitor has circular plates each of radius 2 cm. The plates are being charged such that the electric field in the gap changes at a constant rate of 5 × 10¹¹ V/(m·s). Find: (a) The displacement current (b) The magnetic field at a distance of 1 cm from the axis between the plates (ε₀ = 8.85 × 10⁻¹² F/m, μ₀ = 4π × 10⁻⁷ T·m/A)

Solution

Given:

• R = 2 cm = 0.02 m
• r = 1 cm = 0.01 m (point where B is needed)
• dE/dt = 5 × 10¹¹ V/(m·s)

(a) Displacement current:

Total area: $A = \pi R^2 = \pi \times (0.02)^2 = 1.257 \times 10^{-3}$ m²

$$I_d = \epsilon_0 A \frac{dE}{dt}$$ $$I_d = 8.85 \times 10^{-12} \times 1.257 \times 10^{-3} \times 5 \times 10^{11}$$ $$I_d = 5.56 \text{ A}$$

(b) Magnetic field at r = 1 cm:

Since r < R (inside the capacitor region), we use Ampere-Maxwell law on a circular path of radius r:

Displacement current through circle of radius r:

$$I_{d,enclosed} = \epsilon_0 (\pi r^2) \frac{dE}{dt}$$ $$I_{d,enclosed} = 8.85 \times 10^{-12} \times \pi \times (0.01)^2 \times 5 \times 10^{11}$$ $$I_{d,enclosed} = 1.39 \text{ A}$$

Applying Ampere-Maxwell law:

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enclosed}$$ $$B \times 2\pi r = \mu_0 I_{d,enclosed}$$ $$B = \frac{\mu_0 I_{d,enclosed}}{2\pi r} = \frac{4\pi \times 10^{-7} \times 1.39}{2\pi \times 0.01}$$ $$B = \frac{2 \times 10^{-7} \times 1.39}{0.01} = 2.78 \times 10^{-5} \text{ T}$$

Answer:

• (a) $I_d = 5.56$ A
• (b) $B = 2.78 \times 10^{-5}$ T or 27.8 μT

Problem 4: Show that for a parallel plate capacitor being charged, the displacement current in the gap equals the conduction current in the connecting wires.

Solution

Conduction current in wire:

$$I_c = \frac{dq}{dt}$$

Electric field between capacitor plates:

$$E = \frac{\sigma}{\epsilon_0} = \frac{q}{\epsilon_0 A}$$

where A = area of plates

Rate of change of electric field:

$$\frac{dE}{dt} = \frac{1}{\epsilon_0 A} \frac{dq}{dt}$$

Displacement current in gap:

$$I_d = \epsilon_0 A \frac{dE}{dt}$$

Substituting:

$$I_d = \epsilon_0 A \times \frac{1}{\epsilon_0 A} \frac{dq}{dt} = \frac{dq}{dt} = I_c$$

$$\boxed{I_d = I_c}$$

Proved: Displacement current in gap equals conduction current in wires!

Physical Meaning: The “circuit” is complete - current is continuous throughout!

Level 3: JEE Advanced

Problem 5: A parallel plate capacitor with square plates of side ‘a’ is being charged. At an instant, the conduction current is I₀. Find the magnitude of displacement current through a circular area of radius r (where r < a/2) parallel to the plates and situated symmetrically between them.

Solution

Since $I_d = I_c$ for the entire capacitor:

Total displacement current through entire capacitor = $I_0$

Area of square plates: $A_{total} = a^2$

Area of circular region: $A_{circle} = \pi r^2$

Since displacement current density is uniform:

$$J_d = \frac{I_0}{a^2}$$

Displacement current through circular area:

$$I_{d,circle} = J_d \times A_{circle} = \frac{I_0}{a^2} \times \pi r^2$$

$$I_{d,circle} = \frac{\pi r^2 I_0}{a^2}$$

Answer: $I_d = \frac{\pi r^2 I_0}{a^2}$

Note: This is proportional to the fraction of area: $\frac{\pi r^2}{a^2}$

Problem 6: Derive the expression for the magnetic field at a distance r from the axis (where r < R) inside a charging capacitor with circular plates of radius R. The electric field increases uniformly at rate dE/dt.

Solution

Setup: Use Ampere-Maxwell law on a circular loop of radius r (r < R) centered on the axis.

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enclosed}$$

Left side: By symmetry, B is constant on the circle and tangent to it:

$$\oint \vec{B} \cdot d\vec{l} = B \times 2\pi r$$

Right side: Displacement current through area πr²:

$$I_{d,enclosed} = \epsilon_0 (\pi r^2) \frac{dE}{dt}$$

Equating:

$$B \times 2\pi r = \mu_0 \epsilon_0 \pi r^2 \frac{dE}{dt}$$ $$B = \frac{\mu_0 \epsilon_0 r}{2} \frac{dE}{dt}$$

$$\boxed{B = \frac{\mu_0 \epsilon_0 r}{2} \frac{dE}{dt}}$$

Key observations:

1. B ∝ r (inside the capacitor)
2. B = 0 at center (r = 0)
3. B increases linearly with distance from axis

Vector form: $\vec{B}$ forms concentric circles around the axis, direction given by right-hand rule with displacement current.

Proved

Significance of Displacement Current

1. Completes Maxwell’s Equations

Makes electromagnetic theory consistent and symmetric.

2. Predicts Electromagnetic Waves

Changing E-field creates B-field, which creates E-field, and so on → propagating EM waves!

3. Continuity of Current

Ensures current is continuous even through capacitors and gaps.

4. Practical Applications

• Capacitive circuits in AC
• Wireless power transfer
• Radio wave generation
• Antenna theory

Electromagnetic Wave Generation

From Maxwell’s equations with displacement current:

Wave equation in vacuum:

$$\frac{\partial^2 \vec{E}}{\partial t^2} = c^2 \nabla^2 \vec{E}$$

where $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$ = speed of light

This shows: Electromagnetic waves can propagate through vacuum!

Cross-Links to Other Topics

Related to Electromagnetism

• Ampere’s Law - Original formulation
• Faraday’s Law - Changing magnetic field
• Capacitors - Charging and discharging

Related to EM Waves

• EM Wave Properties - Wave propagation
• EM Spectrum - Different wavelengths

Related to Optics

• Light as EM Wave - Electromagnetic nature
• Polarization - Transverse wave property

Key Takeaways

1. Displacement current is due to changing electric field, not charge flow
2. $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ - fundamental formula
3. In capacitors: $I_d = I_c$ (displacement equals conduction)
4. Maxwell’s addition makes Ampere’s law work for time-varying fields
5. No displacement current ⇒ No electromagnetic waves!
6. Displacement current produces magnetic field just like conduction current
7. Critical for EM wave theory and modern technology

Exam Tips

• JEE Main: Focus on calculating $I_d$ in capacitors; know $I_d = I_c$ relation
• JEE Advanced: Expect derivations of B-field using Ampere-Maxwell law, conceptual questions
• Common trap: Thinking displacement current involves charge flow (it doesn’t!)
• Formula recall: $I_d = \epsilon_0 A \frac{dE}{dt}$ for uniform field
• Symmetry: Changing E → B (Maxwell) like changing B → E (Faraday)
• Conceptual: Understand WHY original Ampere’s law failed (capacitor paradox)
• Units: Check that $I_d$ comes out in Amperes

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Last updated: March 5, 2025