PhysicsISC Class 12 · Electromagnetic Waves

Displacement Current: Maxwell's Correction to Ampère's Law

Updated 15 October 2026 · Fact-checked

Displacement current is the current-like term produced by a changing electric field, $I_d = \varepsilon_0\frac{d\Phi_E}{dt}$. Maxwell added it to Ampère's law so that the law also works in the gap of a charging capacitor. Between the plates it equals the conduction current in the wires, and it produces a magnetic field just as a real current does.

The pot surface meets no wire, so Maxwell's displacement current Id = ε₀dΦ/dt in the gap fills in, equalling i and fixing Ampère's law.

Illustration drawn for QuestionBuddy.

Understand Displacement Current

The problem with Ampère's law. Ampère's circuital law says $\oint B \cdot d l = μ_{0} I$ , where I is the current through any surface bounded by the closed loop. Take a capacitor being charged by a current i, and a circular loop round the wire just outside one plate.

Surface 1, a flat disc spanning the loop, is pierced by the wire: enclosed current = i, so B ≠ 0.
Surface 2, a "pot" with the same rim that bulges into the gap between the plates, is pierced by no wire: enclosed current = 0, so B = 0.

The same loop cannot have two different values of $\oint B \cdot d l$ . Ampère's law is therefore incomplete whenever charge is piling up somewhere.

What is happening in the gap. While the capacitor charges, the charge q on the plates grows, so the electric field between them, $E = \frac{q}{ε _{0} A}$ , grows too. The electric flux through surface 2 is $Φ_{E} = E A = \frac{q}{ε _{0}}$ . Differentiating:
$ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i .$
So the rate of change of electric flux, multiplied by $ε_{0}$ , has the units of current and equals the conduction current in the wires.

Maxwell's idea. Maxwell called this quantity the displacement current,
$I_{d} = ε_{0} \frac{d Φ _{E}}{d t},$
and added it to Ampère's law. The Ampère-Maxwell law is
$\oint B \cdot d l = μ_{0} (I_{c} + ε_{0} \frac{d Φ _{E}}{d t}) = μ_{0} (I_{c} + I_{d}) .$
Through surface 1 only $I_{c}$ contributes; through surface 2 only $I_{d}$ contributes; both give $μ_{0} i$ . The inconsistency disappears and "current" is continuous round the circuit: conduction current in the wires, displacement current in the gap.

Key points.

$I_{d}$ is not a flow of charge; it is a changing electric field.
It exists only while E changes. When the capacitor is fully charged on d.c., dE/dt = 0 and $I_{d}$ = 0.
In terms of the circuit, $I_{d} = C \frac{d V}{d t}$ for a capacitor (since q = CV).
It produces a magnetic field in the same way as conduction current; in the gap of circular plates the field circles the axis.

Why it matters. Faraday's law says a changing B produces E. The displacement current term says a changing E produces B. Together they let changing electric and magnetic fields sustain each other in empty space, travelling as an electromagnetic wave. Maxwell's prediction of such waves, at speed $\frac{1}{μ _{0} ε _{0}}$ , was confirmed by Hertz. The syllabus asks only for the basic idea; no wave equation is needed.

Displacement Current: formulas and key facts

Displacement current: $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$; Ampere.
Parallel plates: $I_{d} = ε_{0} A \frac{d E}{d t}$; Field uniform between plates.
Circuit form: $I_{d} = C \frac{d V}{d t} = \frac{d q}{d t}$; Equals the charging current.
Ampère-Maxwell law: $\oint B \cdot d l = μ_{0} (I_{c} + I_{d})$; Valid for any surface.
Field near the gap (beyond the plate edge): $B = \frac{μ _{0} I}{2 π r}$; Same as near a wire.

How to solve Displacement Current questions

Decide whether the data are about fields (E, flux) or the circuit (C, V, i); both routes give the same current.

1If the charging current i is given: $I_{d} = i$ in the gap.
2If $\frac{d Φ _{E}}{d t}$ is given: $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ with $ε_{0} = 8.854 \times 1 0^{- 12}$ C² N⁻¹ m⁻².
3If $\frac{d E}{d t}$ and plate area are given: $I_{d} = ε_{0} A \frac{d E}{d t}$ .
4If C and $\frac{d V}{d t}$ are given: $I_{d} = C \frac{d V}{d t}$ .
5For B, apply the Ampère-Maxwell law to a circle of radius r: $B \cdot 2 π r = μ_{0} \times$ (current enclosed, displacement or conduction).

Displacement Current in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A camera flash charging

Suppose the 100 µF capacitor of a camera flash is charging and its voltage is rising at 3000 V/s at one instant.

Find the displacement current between its plates at that instant.

Try it first, then show the solutionSolution

$I_{d} = C \frac{d V}{d t} = 100 \times 1 0^{- 6} \times 3000 = 0.3$ A.
This equals the conduction current in the leads at the same instant.

Answer: 0.3 A, the same as in the wires.

Memory hook: The wire's current keeps going across the gap, but as a changing field.

Signals through empty space

Imagine sending a radio message to a spacecraft far from the Earth, where there are no wires and no charges between the antennas.

Explain how a magnetic field can exist in that empty space.

Try it first, then show the solutionSolution

The changing magnetic field of the wave induces a changing electric field (Faraday).
That changing electric field is a displacement current, which creates a magnetic field (Ampère-Maxwell).
Each keeps creating the other, so the wave travels on with no current anywhere.

Answer: A changing E field acts as a source of B, so the fields sustain each other in vacuum.

Memory hook: Changing B makes E; changing E makes B: the wave pulls itself along.

Tricks to remember Displacement Current

Short memory aids. Say each one aloud once, then use it on a question the same day.

Epsilon times flux rate
$I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ : "E-not, d-phi by d-t".
Use it when: Formula recall.
Two surfaces, one answer
Disc gets $I_{c}$ , pot gets $I_{d}$ ; Maxwell makes them equal.
Use it when: Explaining the inconsistency.
No change, no displacement
Steady E (charged capacitor on d.c.) gives $I_{d}$ = 0.
Use it when: Trick questions.
Faraday mirrored
Changing B → E (Faraday); changing E → B (Maxwell).
Use it when: Linking to EM waves.

Displacement Current: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Source of displacement current · 1 mark · MCQBoard level

Displacement current is produced by:
(a) a steady electric field
(b) a changing electric field
(c) a steady magnetic field
(d) moving electrons in a conductor

Try it first, then show the solutionSolution

$I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ , which is non-zero only when the electric flux changes.

Answer: (b).

Example 2: Current and flux rate in the gap · 2 marks · Short answerBoard level

A capacitor is being charged by a current of 0.8 A. Find the displacement current between its plates and the rate of change of electric flux there.

Try it first, then show the solutionSolution

Displacement current in the gap = conduction current = 0.8 A. (1)
$\frac{d Φ _{E}}{d t} = \frac{I _{d}}{ε _{0}} = \frac{0.8}{8.854 \times 1 0 ^{- 12}} \approx 9.04 \times 1 0^{10}$ V m s⁻¹. (1)

Answer: 0.8 A; ≈ $9.04 \times 1 0^{10}$ V m s⁻¹.

Example 3: Why Maxwell needed it · 3 marks · Long answerHarder board level

Using a capacitor being charged, explain the inconsistency in Ampère's circuital law and how Maxwell removed it. Write the Ampère-Maxwell law.

Try it first, then show the solutionSolution

For a loop round the wire, a flat surface cut by the wire encloses current i, but a surface with the same rim passing through the gap encloses no conduction current; Ampère's law gives two different values of $\oint B \cdot d l$ . (1)
In the gap the electric flux $Φ_{E} = \frac{q}{ε _{0}}$ changes, and $ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i$ . Maxwell called this the displacement current. (1)
$\oint B \cdot d l = μ_{0} (I_{c} + ε_{0} \frac{d Φ _{E}}{d t})$ ; both surfaces now give $μ_{0} i$ . (1)

Answer: Adding $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ makes the law hold for every surface.

Example 4: Field route and circuit route · 3 marks · Short answerHarder board level

A 5 nF parallel-plate capacitor with plate area 0.05 m² has its voltage rising at $2 \times 1 0^{6}$ V/s. Find the displacement current (a) using $C \frac{d V}{d t}$ and (b) using $ε_{0} A \frac{d E}{d t}$ , after finding the plate separation.

Try it first, then show the solutionSolution

(a) $I_{d} = 5 \times 1 0^{- 9} \times 2 \times 1 0^{6} = 0.01$ A = 10 mA. (1)
(b) $d = \frac{ε _{0} A}{C} = \frac{8.854 \times 1 0 ^{- 12} \times 0.05}{5 \times 1 0 ^{- 9}} \approx 8.85 \times 1 0^{- 5}$ m; $\frac{d E}{d t} = \frac{1}{d} \frac{d V}{d t}$ . (1)
$I_{d} = ε_{0} A \times \frac{2 \times 1 0 ^{6}}{8.854 \times 1 0 ^{- 5}} = 0.01$ A, the same. (1)

Answer: 10 mA by both routes.

Example 5: Magnetic field inside and outside the gap · 3 marks · Short answerHOTS / competency

A capacitor with circular plates of radius 0.1 m is charged by a current of 1.5 A. Find the magnetic field (a) in the gap at 0.05 m from the axis, and (b) at 0.2 m from the axis.

Try it first, then show the solutionSolution

(a) Displacement current is spread uniformly over the plate area; a circle of radius 0.05 m encloses $1.5 \times (\frac{0.05}{0.1})^{2} = 0.375$ A. (1)
$B = \frac{μ _{0} \times 0.375}{2 π \times 0.05} = 1.5 \times 1 0^{- 6}$ T. (1)
(b) Beyond the plates the whole 1.5 A is enclosed: $B = \frac{μ _{0} \times 1.5}{2 π \times 0.2} = 1.5 \times 1 0^{- 6}$ T. (1)

Answer: Both $1.5 \times 1 0^{- 6}$ T (inside, B grows in proportion to r; outside, it falls as $\frac{1}{r}$ ).

Example 6: Capacitor on a.c. · 4 marks · Case-basedHOTS / competency

Suppose a 10 µF capacitor is connected to $V = 200 sin (100 π t)$ V.
(i) Find the maximum rate of change of voltage. (1 mark)
(ii) Find the peak displacement current in the gap. (1 mark)
(iii) What is the peak current in the connecting wires? Why? (1 mark)
(iv) Show that the current leads the voltage by 90°. (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{d V}{d t} = 200 \times 100 π cos (100 π t)$ ; maximum $= 2 π \times 1 0^{4} \approx 6.28 \times 1 0^{4}$ V/s.
(ii) $I_{d} = C \frac{d V}{d t}$ , peak $= 1 0^{- 5} \times 6.28 \times 1 0^{4} \approx 0.628$ A.
(iii) 0.628 A: the conduction current in the wires equals the displacement current in the gap at every instant.
(iv) $I = 0.628 cos (100 π t) = 0.628 sin (100 π t + \frac{π}{2})$ : the current leads by 90°.

Answer: (i) ≈ $6.28 \times 1 0^{4}$ V/s; (ii) ≈ 0.628 A; (iii) 0.628 A, continuity of current; (iv) cos form shows a 90° lead.

Where students lose marks in Displacement Current

The slips examiners see most often, and what to write instead.

Mistake: Calling displacement current a flow of charge
No charge crosses the gap of a capacitor.
Do this: Say it is a changing electric field that has the same magnetic effect as a current.
Usually costs: 1 mark.
Mistake: Leaving out ε₀
Writing $I_{d} = \frac{d Φ _{E}}{d t}$ gives the wrong units.
Do this: $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ .
Usually costs: 1 mark.
Mistake: Describing only one surface
The inconsistency needs two surfaces with the same boundary; one surface does not show it.
Do this: Name both: the flat disc (current i) and the pot through the gap (current 0).
Usually costs: 1 mark.
Mistake: Saying $I_{d}$ exists for a charged capacitor on d.c.
Students think the gap always carries current.
Do this: $I_{d}$ needs a changing field; on steady d.c. it is zero.
Usually costs: 1 mark.

Exam tips

Draw the capacitor, the loop and the two surfaces in an explanation question.
Write the Ampère-Maxwell law with both terms in brackets.
State that $I_{d}$ equals the conduction current in the leads at every instant.
Use $ε_{0} = 8.854 \times 1 0^{- 12}$ C² N⁻¹ m⁻² unless the paper gives another value.

Practice questions on Displacement Current

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of displacement current is:
(a) coulomb
(b) volt metre
(c) ampere
(d) tesla
Hint
It equals the conduction current.
Answer
(c).
Q2 · 1 mark · Short answerBoard level
The electric flux through a region changes at $4.5 \times 1 0^{11}$ V m s⁻¹. Find the displacement current.
Hint
Multiply by ε₀.
Answer
≈ 3.98 A.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A magnetic field exists in the gap of a capacitor while it is being charged.
Reason (R): A changing electric field produces a magnetic field.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Ampère-Maxwell law.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
The plates of a capacitor have area 0.02 m² and the field between them grows at $5 \times 1 0^{11}$ V m⁻¹ s⁻¹. Find the displacement current.
Hint
$I_{d} = ε_{0} A \frac{d E}{d t}$ .
Answer
≈ 0.0885 A (88.5 mA).
Q5 · 1 mark · Short answerHarder board level
A capacitor is connected to a battery and has become fully charged. What is the displacement current between its plates now? Explain.
Hint
Is E changing?
Answer
Zero: the field between the plates is now steady, so $\frac{d Φ _{E}}{d t} = 0$ .
Q6 · 2 marks · Short answerHOTS / competency
The voltage across a 2 µF capacitor rises uniformly from 0 to 100 V in 4 ms. Find the displacement current during charging.
Hint
$\frac{d V}{d t} = 2.5 \times 1 0^{4}$ V/s.
Answer
0.05 A.
Q7 · 2 marks · Short answerHOTS / competency
Explain, using the Ampère-Maxwell law, why the idea of displacement current led Maxwell to predict electromagnetic waves.
Hint
Changing E makes B; changing B makes E.
Answer
Faraday's law shows a changing B creates E; the displacement current term shows a changing E creates B. Each sustains the other, so a disturbance in the fields can travel through empty space as a wave, at $\frac{1}{μ _{0} ε _{0}} = c$ .
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a capacitor has circular plates of radius 5 cm, 2 mm apart, and is charged by a current of 0.2 A.
(i) What is the displacement current in the gap? (1 mark)
(ii) Find the rate of change of the electric field between the plates. (1 mark)
(iii) Find B at 5 cm from the axis, at the edge of the plates. (1 mark)
(iv) Find B at 2.5 cm from the axis, inside the gap. (1 mark)
Hint
$A = π (0.05)^{2}$ .
Answer
(i) 0.2 A. (ii) $\frac{d E}{d t} = \frac{I}{ε _{0} A} \approx 2.88 \times 1 0^{12}$ V m⁻¹ s⁻¹. (iii) $B = \frac{μ _{0} \times 0.2}{2 π \times 0.05} = 8 \times 1 0^{- 7}$ T. (iv) Enclosed $I_{d}$ = 0.05 A, so $B = 4 \times 1 0^{- 7}$ T.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Formula for displacement current?Show answer
$I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ .
2.Ampère-Maxwell law?Show answer
$\oint B \cdot d l = μ_{0} (I_{c} + I_{d})$ .
3.Charging current 2 A: displacement current in the gap?Show answer
2 A.
4.Who introduced displacement current?Show answer
James Clerk Maxwell.
5.Displacement current in a fully charged capacitor on d.c.?Show answer
Zero.

Displacement Current: frequently asked questions

Is displacement current a real current?

It is not a flow of charge, but it produces a magnetic field exactly as a conduction current does, so it is real in its magnetic effect.

Why is it called "displacement" current?

Maxwell first pictured it as a small shift (displacement) of charges in the medium between the plates. The name stayed even though it exists in vacuum too.

Where does displacement current occur besides capacitors?

Anywhere an electric field changes with time, for example in the space through which an electromagnetic wave passes.

What does the ISC 2027 syllabus need here?

The basic idea and concept of displacement current, as the source of a magnetic field due to a changing electric field and the reason electromagnetic waves exist; descriptions are qualitative.

How is this likely to be asked?

As a short explanation (why Maxwell added it, or its expression) or a one-step numerical relating it to the charging current, \frac{d\Phi_E}{dt} or C\frac{dV}{dt}.