Displacement current — Maxwell's addition to Ampere's law: Question bank

Before you start, three anchors we will lean on repeatedly:

Conduction current $I_{c}$ = actual charge moving (electrons drifting through a wire).
Displacement current $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ = the magnetic-field-making effect of a changing electric flux — no charge moves.
The whole law: $\oint B \cdot d l = μ_{0} (I_{c} + I_{d})$ from Ampère's Law as corrected by Maxwell.

True or false — justify

A "changing electric field is literally a flow of charge"

False — nothing physical moves in the gap; charge only piles onto the plates.

I_{d}

has the units and magnetic effect of a current but is not a current of matter.

Displacement current only exists inside capacitors

False — it exists anywhere the electric flux changes, including empty vacuum during an EM wave, where there are no plates and no charges at all.

In the wire (not the gap), the displacement current is zero

Mostly true for an ideal wire: inside a good conductor

E \approx 0

, so

d Φ_{E} / d t \approx 0

and

I_{d} \approx 0

; the conduction current

I_{c}

carries everything there.

Removing

I_{d}

from Ampère's law would still leave a self-consistent theory, just without waves

False — without

I_{d}

the law is outright contradictory at a capacitor (two surfaces, two answers). It is not merely incomplete; it is broken.

Displacement current requires a magnetic field to already be present

False —

I_{d}

is caused by a changing

E

and it is what creates the

B

. Cause is the changing flux; the

B

loop is the effect.

For a steady (constant) current charging nothing — e.g. a plain resistor circuit —

I_{d}

between any two points is zero

True when fields are static: constant

E

means

d Φ_{E} / d t = 0

, so

I_{d} = 0

and ordinary Ampère's Law already works perfectly.

The displacement current in the gap points in the opposite direction to the conduction current in the wire

False — it points the same way and has the same magnitude, which is exactly why the magnetic field is continuous across the gap.

I_{d}

can be nonzero even while the total charge in a region is zero

True — in a passing EM wave the flux through a surface oscillates through zero yet its rate of change

d Φ_{E} / d t

is largest right there, giving a big

I_{d}

with no net charge.

Spot the error

"No charge crosses the gap, therefore no magnetic field exists near the gap."

Error: the conclusion ignores

I_{d}

. A compass near the gap does deflect because the growing

E

supplies

μ_{0} I_{d}

exactly equal to

μ_{0} I_{c}

from the wire.

I_{d} = ε_{0} d Φ_{E} / d t

, and flux

Φ_{E} = E \cdot A

always, so

I_{d} = ε_{0} A d E / d t

in every situation."

Error:

Φ_{E} = E A

only holds when

E

is uniform and perpendicular to

A

. In general

Φ_{E} = \int E \cdot d A

; you cannot pull

E

out of the integral for a fringing or tilted field.

"Ampère's law fails because we chose a stupid bulging surface; use the flat disc and the problem disappears."

Error: a valid law must give one answer for every surface on the loop. That the flat disc "works" does not save a law that fails on the equally-valid bulging one.

"Since

I_{d} = I_{c}

for a capacitor, displacement current is just a fancy renaming of conduction current."

Error: they are equal in this special case by charge conservation, not identical concepts. In a vacuum wave

I_{c} = 0

but

I_{d} \neq = 0

— the renaming would make no sense there.

"The displacement current density is

J_{d} = ε_{0} E

Error: it is the time derivative of the field,

J_{d} = ε_{0} \partial E / \partial t

. A large but constant

E

gives zero

J_{d}

"Between the plates

E = σ / ε_{0}

, so

Φ_{E}

depends on the plate separation

d

Error: for the ideal parallel-plate field,

Φ_{E} = E A = Q / ε_{0}

depends only on the charge, not on

d

— the separation cancels out (see Parallel Plate Capacitor).

Why questions

Why did Maxwell add a term to

E

's time-derivative and not, say, to

B

Because the broken law was Ampère's (

B

circulation vs current); Faraday's Law of Induction already had the changing-

B

term. Maxwell supplied the missing mirror term so a changing

E

makes

B

Why is the constant out front exactly

ε_{0}

and not some other value?

It is forced by matching:

Φ_{E} = Q / ε_{0}

from Gauss's Law, so

ε_{0} d Φ_{E} / d t = d Q / d t = I

. Any other constant would break the equality

I_{d} = I_{c}

Why does the corrected law suddenly predict a speed,

c = 1/ μ_{0} ε_{0}

Because

E

and

B

now feed each other through

μ_{0} ε_{0} \partial E / \partial t

; combining that with Faraday's term yields a wave equation whose speed is fixed by those two constants — the Speed of Light.

Why must the same

\oint B \cdot d l

hold for every surface on a loop?

The left side depends only on the loop (the boundary), so the right side must too. If it changed with the surface, the equation would define no single number and be meaningless.

Why does

I_{d}

complete Maxwell's Equations rather than just patch one of them?

It restores the symmetry between electric and magnetic induction, closing the feedback loop that makes the four equations self-consistent and wave-supporting.

Why is

I_{d}

inside a wire negligible but inside the gap huge?

In the conductor

E \approx 0

(charges rearrange to cancel it), so its rate of change is tiny; across the gap the whole voltage sits, so

E

and its growth are large.

Edge cases

What is

I_{d}

the instant the capacitor is fully charged (current stops)?

Zero — charging halts, so

d Q / d t = 0

, hence

d Φ_{E} / d t = 0

and

I_{d} = 0

. The magnetic field between the plates also vanishes at that instant.

What is the magnetic field exactly on the central axis (

r = 0

) between charging plates?

Zero, by symmetry:

B = μ_{0} I r / (2 π R^{2})

gives

B (0) = 0

, just like on the axis of a uniform current-carrying wire.

What happens to

I_{d}

if the capacitor discharges instead of charges?

It reverses sign along with

d Q / d t

;

E

now shrinks, so

d Φ_{E} / d t < 0

, and the induced

B

circulates the opposite way — consistent with the reversed wire current.

Just outside the plates (at radius

r > R

), does the formula

B = μ_{0} I r / (2 π R^{2})

still hold?

No — beyond the plate edge the entire displacement current

I_{d} = I

is enclosed, so

B = μ_{0} I / (2 π r)

, identical to the field around the feeding wire. The linear-in-

r

form is only for

r < R

If the electric field grows but stays perfectly uniform and infinite in extent, is there still a

B

A changing uniform

E

does define a nonzero

J_{d}

, but by symmetry an infinite uniform source produces no net circulating

B

— the geometry, not just

I_{d}

, fixes the field.

In a region of pure vacuum with light passing through, what carries the "current" in Ampère–Maxwell?

Entirely

I_{d}

I_{c} = 0

because there are no charges, so

\oint B \cdot d l = μ_{0} ε_{0} d Φ_{E} / d t

alone sustains the wave.

At the moment an EM wave's

E

passes through zero, is

I_{d}

also zero there?

No —

E = 0

but its slope in time is steepest right then, so

d Φ_{E} / d t

(hence

I_{d}

and the resulting

B

) is at its maximum. Field value and its rate of change are 90° out of phase.

Connections

Ampère's Law — the law these traps stress-test
Faraday's Law of Induction — the mirror term
Maxwell's Equations — the completed set
Electromagnetic Waves — where $I_{c} = 0$ but $I_{d} \neq = 0$
Parallel Plate Capacitor — the canonical geometry
Gauss's Law — supplies $Φ_{E} = Q / ε_{0}$
Speed of Light — the emergent constant $1/ μ_{0} ε_{0}$