JEE Mains Physics · Electromagnetic Waves

Displacement Current, Maxwell's Equations and Wave Speed

A changing electric field acts like a current, and with it Maxwell's four equations predict waves that travel at 1/√(με): c in vacuum, c/n in a medium.

All Electromagnetic Waves notes JEE Mains Physics strategy

Why this matters

Twenty-eight PYQs, seven of them asking for a number, and six from 2026. Seven are about the displacement current between capacitor plates. Six match Maxwell's four equations to their names or ask what can produce a changing field. Fifteen find the speed of the wave in a medium, from its relative permittivity and permeability or from the phase of the field. That last group is nearly always one line: read v from the phase, then n = c/v.

Concept 1 of 3: Displacement current in a capacitor

While a capacitor charges, current flows in the wires but no charge crosses the gap. Ampere's law would then give a magnetic field around the wire and none around the gap. Maxwell fixed this by counting a changing electric flux as a current, the displacement current. Between the plates it is exactly equal to the current in the leads.

Definition

Displacement current: $i_d = \varepsilon_0\,\dfrac{d\Phi_E}{dt}$ . It has the dimensions of current.
Between capacitor plates $\Phi_E = EA = q/\varepsilon_0$ , so $i_d = \dfrac{dq}{dt} = C\dfrac{dV}{dt}$ , equal to the conduction current in the leads at every instant. With an AC supply the rms values are equal too.
For a parallel-plate capacitor $C = \varepsilon_0A/d$ , so $i_d = \dfrac{\varepsilon_0A}{d}\dfrac{dV}{dt} = \varepsilon_0A\dfrac{dE}{dt}$ .
The field between the plates is uniform, so a surface of area $A_0$ inside the gap, parallel to the plates, carries $i_d\,A_0/A$ .
On an AC supply the current is $I_{rms} = V_{rms}/X_C = V_{rms}\,\omega C$ .
In a conducting medium with $E = E_0\sin\omega t$ : $j_c = \sigma E$ and $j_d = \varepsilon_0\,\partial E/\partial t$ , so the ratio of their peaks is $\sigma/(\varepsilon_0\omega)$ .

Displacement current

i_d = \varepsilon_0\frac{d\Phi_E}{dt} = C\frac{dV}{dt}, \qquad \frac{(j_c)_0}{(j_d)_0} = \frac{\sigma}{\varepsilon_0\omega}

Worked example

An air capacitor has plates of area 0.02 m² that are 1 mm apart. The voltage across it rises at

5 \times 10^{5}

V/s. Find (a) its capacitance, (b) the displacement current between the plates, and (c) the displacement current through a 50 cm² surface inside the gap, parallel to the plates. (

\varepsilon_0 = 8.85 \times 10^{-12}

F/m)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q8Moderate

Example 1 · Electromagnetic Waves · Displacement Current, Maxwell's Equations and Wave Speed

A parallel plate capacitor has a capacitance

C = 200pF

. It is connected to

230\text{ }V

ac supply with an angular frequency

300rad/s

. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are:

The displacement current equals the conduction current

Between the plates of a capacitor the displacement current is exactly the current flowing in the leads, at every instant and in rms value. An option that makes one of them ten times the other is wrong.

A surface that covers part of the gap takes its share

The field between the plates is uniform, so a surface of area A₀ inside the gap carries only the fraction A₀/A of the displacement current, not all of it.

Use ω, not f, in σ/ε₀ω

The displacement current density peaks at ε₀ωE₀, with ω = 2πf. Writing f in place of ω makes the ratio of conduction to displacement current 2π times too small.

Drill 6 more on displacement current in a capacitor

Concept 2 of 3: Maxwell's four equations and their names

Two of the equations are about the flux through a closed surface: they say where field lines start and end. The other two are about the circulation around a loop: they say that a changing field of one kind makes a field of the other kind. Together they predict electromagnetic waves.

Definition

Gauss's law for electricity: $\oint \vec E\cdot d\vec A = q/\varepsilon_0$ . Charges are the sources of E.
Gauss's law for magnetism: $\oint \vec B\cdot d\vec A = 0$ . There are no isolated magnetic poles.
Faraday's law: $\oint \vec E\cdot d\vec l = -\,d\Phi_B/dt$ . A changing magnetic flux makes an electric field around a loop.
Ampere-Maxwell law: $\oint \vec B\cdot d\vec l = \mu_0\left(i_c + \varepsilon_0\,d\Phi_E/dt\right)$ . Plain Ampere's law, $\oint \vec B\cdot d\vec l = \mu_0 I$ , holds only for steady currents.
A changing magnetic field comes from a changing current or an accelerated charge. A permanent magnet, a steady current and an electric field that changes at a steady rate all give a magnetic field that does not change in time.
Electromagnetic waves are produced by accelerated charges. A charge at rest or moving at constant velocity does not radiate.

Law	Equation	What it says
Gauss's law for electricity	$\oint \vec E\cdot d\vec A = q/\varepsilon_0$	The electric flux out of a closed surface is the enclosed charge divided by ε₀.
Gauss's law for magnetism	$\oint \vec B\cdot d\vec A = 0$	Magnetic field lines close on themselves; there are no magnetic monopoles.
Faraday's law of induction	$\oint \vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt}$	A changing magnetic flux induces an electric field around a loop.
Ampere-Maxwell law	$\oint \vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$	A conduction current and a changing electric flux both produce a magnetic field.
Ampere's circuital law	$\oint \vec B\cdot d\vec l = \mu_0 I$	The steady-current special case, with no changing electric flux. Without the displacement term it fails across the gap of a charging capacitor.

A closed-surface integral (dA) means a Gauss law; a loop integral (dl) means Faraday or Ampere-Maxwell.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q103Moderate

Example 2 · Electromagnetic Waves · Displacement Current, Maxwell's Equations and Wave Speed

Match List I with List II

LIST I	LIST II
A. Gauss's Law in Electrostatics	I. $\oint\overrightarrow{E}\cdot d\overrightarrow{l} = -\frac{d\phi_{B}}{dt}$
B. Faraday's Law	II. $\oint\overrightarrow{B}\cdot d\overrightarrow{A} = 0$
C. Gauss's Law in Magnetism	III. $\oint\overrightarrow{B}\cdot d\overrightarrow{l} = \mu_{0}i_{c} + \mu_{0}\epsilon_{0}\frac{d\phi_{E}}{dt}$
D. Ampere-Maxwell Law	IV. $\oint\overrightarrow{E}\cdot d\overrightarrow{s} = \frac{q}{\epsilon_{0}}$

Choose the correct answer from the options given below:

Plain Ampere's law is the steady-current law

The circulation of B equals μ₀I only for steady currents. The law that holds when fields change in time carries the extra term μ₀ε₀ dΦ_E/dt.

Read the integral before the symbols

In a match list, a closed-surface integral of E or B is one of the two Gauss laws, and a loop integral is Faraday's law or the Ampere-Maxwell law. Sorting by dA or dl first leaves only two choices to make.

A steady source gives a steady field

A permanent magnet, a direct current and an electric field that grows at a constant rate all produce a magnetic field that does not change with time. A changing magnetic field needs a changing current or an accelerating charge.

Drill 5 more on maxwell's four equations and their names

Concept 3 of 3: Speed of electromagnetic waves in vacuum and in a medium

Maxwell's equations give a wave whose speed depends only on the permeability and permittivity of what it travels through. In vacuum that speed is c. In a medium it drops by √(μᵣεᵣ), which is the refractive index. A question often hides the speed in the phase of the field: v is the ratio of the number in front of t to the number in front of x.

Definition

Vacuum: $c = 1/\sqrt{\mu_0\varepsilon_0} = 3 \times 10^{8}$ m/s.
Medium: $v = 1/\sqrt{\mu\varepsilon} = c/\sqrt{\mu_r\varepsilon_r}$ , and the refractive index is $n = c/v = \sqrt{\mu_r\varepsilon_r}$ .
Non-magnetic medium ( $\mu_r = 1$ ): the dielectric constant is $\varepsilon_r = n^{2}$ .
From the phase: for $E_0\sin(kx - \omega t)$ or $E_0\cos(\omega t - kx)$ , $v = \omega/k$ , $\lambda = 2\pi/k$ and $f = \omega/2\pi$ . In vacuum or air, $\lambda = c/f$ .
From the amplitudes: $v = E_0/B_0$ in any medium.
In a medium the ratio of E to the magnetic intensity H is $\sqrt{\mu/\varepsilon}$ , which is 377 Ω in vacuum.
The frequency does not change when a wave enters a medium; the speed and the wavelength both fall by the factor n.

Wave speed

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}, \qquad v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{\omega}{k}, \qquad n = \sqrt{\mu_r\varepsilon_r}

Worked example

In a non-magnetic medium the electric field of a wave is

E = 30\sin(2.5 \times 10^{7}x - 3 \times 10^{15}t)

V/m. Find the speed of the wave, the refractive index, the dielectric constant, and the wavelength in the medium and in vacuum. (

c = 3 \times 10^{8}

m/s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · 2021 Paper 24 · Q19Moderate

Example 3 · Electromagnetic Waves · Displacement Current, Maxwell's Equations and Wave Speed

Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by

E = 20\cos\left( 2 \times10^{10}t - 200x \right)V/m

. The dielectric constant of the medium is equal to: (Take

\mu_{r}= 1

)

Take the speed from the phase, not c

When ω/k in the phase is not 3 × 10⁸ m/s, the wave is in a medium. Use v = ω/k for the refractive index and for E₀ = vB₀; using c there gives a wrong amplitude.

A magnetic medium needs μᵣ too

The refractive index is √(μᵣεᵣ). If the permeability is given as a multiple of μ₀, it goes under the root with the dielectric constant. Dropping it leaves √εᵣ only.

Square the index, do not double it

For a non-magnetic medium the dielectric constant is n². A wave that travels at half of c has n = 2 and a dielectric constant of 4, not 2.

The frequency stays the same in a medium

Entering a medium changes the speed and the wavelength by the same factor n. The frequency is fixed by the source and does not change.

Drill 14 more on speed of electromagnetic waves in vacuum and in a medium

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Displacement current in a capacitor
Displacement current
$i_d = \varepsilon_0\frac{d\Phi_E}{dt} = C\frac{dV}{dt}, \qquad \frac{(j_c)_0}{(j_d)_0} = \frac{\sigma}{\varepsilon_0\omega}$
Speed of electromagnetic waves in vacuum and in a medium
Wave speed
$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}, \qquad v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{\omega}{k}, \qquad n = \sqrt{\mu_r\varepsilon_r}$

Reference tables (1)

Maxwell's four equations and their names5 rows

Law	Equation	What it says
Gauss's law for electricity	$\oint \vec E\cdot d\vec A = q/\varepsilon_0$	The electric flux out of a closed surface is the enclosed charge divided by ε₀.
Gauss's law for magnetism	$\oint \vec B\cdot d\vec A = 0$	Magnetic field lines close on themselves; there are no magnetic monopoles.
Faraday's law of induction	$\oint \vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt}$	A changing magnetic flux induces an electric field around a loop.
Ampere-Maxwell law	$\oint \vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$	A conduction current and a changing electric flux both produce a magnetic field.
Ampere's circuital law	$\oint \vec B\cdot d\vec l = \mu_0 I$	The steady-current special case, with no changing electric flux. Without the displacement term it fails across the gap of a charging capacitor.

A closed-surface integral (dA) means a Gauss law; a loop integral (dl) means Faraday or Ampere-Maxwell.

Watch out for (10)

The displacement current equals the conduction current
→ Displacement current in a capacitor
A surface that covers part of the gap takes its share
→ Displacement current in a capacitor
Use ω, not f, in σ/ε₀ω
→ Displacement current in a capacitor
Plain Ampere's law is the steady-current law
→ Maxwell's four equations and their names
Read the integral before the symbols
→ Maxwell's four equations and their names
A steady source gives a steady field
→ Maxwell's four equations and their names
Take the speed from the phase, not c
→ Speed of electromagnetic waves in vacuum and in a medium
A magnetic medium needs μᵣ too
→ Speed of electromagnetic waves in vacuum and in a medium
Square the index, do not double it
→ Speed of electromagnetic waves in vacuum and in a medium
The frequency stays the same in a medium
→ Speed of electromagnetic waves in vacuum and in a medium

Test yourself on Electromagnetic Waves

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