PhysicsCBSE Class 12 · Electromagnetic Waves

Displacement Current Class 12: Changing Electric Field as a Current

Updated 15 October 2026 · Fact-checked

Displacement current is the 'current' associated with a changing electric field: i_d = ε₀ dΦ_E/dt. Maxwell added it to Ampere's law because a charging capacitor's gap carries no conduction current yet has a magnetic field around it. Between the plates, the displacement current equals the conduction current in the wires, so current is continuous.

No charge crosses the gap, but the changing E field gives i_d equal to the wire current i_c, so Ampere's law gives the same B for both surfaces.

Illustration drawn for QuestionBuddy.

Understand Displacement Current

A charging capacitor is a puzzle for Ampere's circuital law. Current flows in the wires, yet no charge crosses the gap between the plates. Maxwell's answer was that a changing electric field acts as a source of magnetic field, just as a current does. The 2026-27 syllabus asks for the basic idea of this displacement current.

The inconsistency. Ampere's law says $\oint B \cdot d l = μ_{0} i$ , where i is the current through any surface bounded by the loop. Take a small circular loop round the wire leading to a capacitor plate. A flat surface bounded by the loop is pierced by the wire, so $i = i_{c}$ and B is not zero. But stretch the surface into a balloon that passes between the plates without touching them: no conduction current crosses it, so the law would give B = 0 on the same loop. The same loop cannot have two values of $\oint B \cdot d l$ , so Ampere's law is incomplete.

What is changing between the plates. As charge q builds up on a plate of area A, the field between the plates is $E = \frac{q}{ε _{0} A}$ , and the electric flux through the balloon surface is $Φ_{E} = E A = \frac{q}{ε _{0}}$ . So $ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i_{c}$ : the rate of change of electric flux, multiplied by $ε_{0}$ , equals exactly the conduction current in the wire.

Displacement current. Maxwell defined $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ and wrote the Ampere-Maxwell law: $\oint B \cdot d l = μ_{0} (i_{c} + i_{d}) = μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ _{E}}{d t}$ . Now the flat surface gives $μ_{0} i_{c}$ and the balloon gives $μ_{0} i_{d}$ , which is the same. Key points:

Displacement current is not a flow of charge; it is the effect of a changing electric field.
In the wires, $i_{c}$ flows and $i_{d}$ is negligible; between the plates, $i_{c} = 0$ and $i_{d} = i_{c}$ (of the wire). Total current $i_{c} + i_{d}$ is the same everywhere along the circuit: current is continuous.
It exists only while the field changes. A fully charged capacitor on steady DC has $i_{d} = 0$ .
In an AC circuit with a capacitor, the displacement current between the plates alternates along with the conduction current.

Why it matters. Faraday's law says a changing magnetic field produces an electric field. Maxwell's term says a changing electric field produces a magnetic field. Together they allow electric and magnetic fields to keep regenerating each other and travel through empty space as electromagnetic waves, at speed $c = \frac{1}{μ _{0} ε _{0}}$ . The magnetic fields from displacement currents in ordinary circuits are tiny, which is why the effect went unnoticed before Maxwell.

Displacement Current: formulas and key facts

Displacement current: $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$; Φ_E = electric flux; unit of i_d: ampere.
Ampere-Maxwell law: $\oint B \cdot d l = μ_{0} (i_{c} + i_{d})$; Total current is continuous.
Parallel plate capacitor: $i_{d} = ε_{0} A \frac{d E}{d t} = C \frac{d V}{d t} = i_{c}$; Displacement current in the gap equals conduction current in the wires.
Field inside circular plates (radius R): $B = \frac{μ _{0} i _{d} r}{2 π R ^{2}} (r < R), B = \frac{μ _{0} i _{d}}{2 π r} (r > R)$; From the Ampere-Maxwell law with a circular loop.
Speed of EM waves: $c = \frac{1}{μ _{0} ε _{0}} \approx 3 \times 1 0^{8} m/s$; Displacement current makes EM waves possible.

How to solve Displacement Current questions

Use this method for displacement current questions.

1If the conduction current in the wires is known, the displacement current between the plates is the same.
2If the rate of change of field is given: $i_{d} = ε_{0} A \frac{d E}{d t}$ ; of flux: $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ ; of voltage: $i_{d} = C \frac{d V}{d t}$ .
3For the magnetic field between circular plates, use a circular loop of radius r: inside, only the fraction $\frac{r ^{2}}{R ^{2}}$ of $i_{d}$ is enclosed.
4For a steady (fully charged) DC situation, the field is not changing: $i_{d} = 0$ .
5Give units: A for current, V m⁻¹ s⁻¹ for dE/dt, T for B.

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.

Displacement current in a charging capacitor

Suppose a camera flash capacitor made of rolled metal foils, with an effective plate area of 0.5 m², is being charged with a steady current of 2 A.

(a) What is the displacement current between the plates? (b) At what rate is the electric field between the plates changing? (c) What happens to the displacement current once the capacitor is fully charged?

Try it first, then show the solutionSolution

Displacement current between the plates equals the conduction current in the leads: 2 A.
$\frac{d E}{d t} = \frac{i _{d}}{ε _{0} A} = \frac{2}{8.854 \times 1 0 ^{- 12} \times 0.5} \approx 4.5 \times 1 0^{11}$ V m⁻¹ s⁻¹.
Once fully charged, E stops changing, so $i_{d} = 0$ (and the conduction current is zero too).

Answer: (a) 2 A; (b) about $4.5 \times 1 0^{11}$ V m⁻¹ s⁻¹; (c) it becomes zero.

Memory hook: Charge flows to the plates; field change carries the current across the gap.

Why the effect went unnoticed

Imagine trying to detect the magnetic field produced by a displacement current of 1 A between the plates of a capacitor, at a point 5 cm from the axis, outside the edges of the plates.

(a) Estimate the field there. (b) Compare it qualitatively with the field near a magnet, and say why Ampere could miss this effect.

Try it first, then show the solutionSolution

Outside the plates the loop encloses the whole displacement current: $B = \frac{μ _{0} i _{d}}{2 π r} = \frac{4 π \times 1 0 ^{- 7} \times 1}{2 π \times 0.05} = 4 \times 1 0^{- 6}$ T.
This is only a few microtesla, thousands of times weaker than the field near a small magnet, and it lasts only while the capacitor charges. In everyday circuits such fields are hard to notice without sensitive instruments.

Answer: (a) about $4 \times 1 0^{- 6}$ T; (b) tiny and short-lived, so easily missed.

Memory hook: Displacement current is real but quiet until it becomes a wave.

Tricks to remember Displacement Current

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

Changing E is a current
i_d = ε₀ × (rate of change of electric flux): 'epsilon-nought d-phi-E by d-t'.
Use it when: Recalling the formula.
Mirror of Faraday
Faraday: changing B makes E. Maxwell: changing E makes B.
Use it when: Explaining why EM waves exist.
Wire carries charge, gap carries change
i_c in the wire equals i_d in the gap.
Use it when: Continuity of current.
No change, no i_d
Steady field, fully charged capacitor: zero displacement current.
Use it when: DC questions.

Displacement Current: solved examples

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

Example 1: Displacement current equals conduction current · 1 mark · Very short answerBoard level

A parallel plate capacitor is being charged by a current of 0.5 A. (a) What is the displacement current between its plates? (b) What is the conduction current between the plates?

Try it first, then show the solutionSolution

(a) $i_{d} = i_{c} = 0.5$ A; (b) zero, since no charge crosses the gap. (1)

Answer: (a) 0.5 A; (b) zero.

Example 2: Need for displacement current · 3 marks · Short answerBoard level

Explain why Ampere's circuital law is inconsistent for a charging capacitor, and show that $ε_{0} \frac{d Φ _{E}}{d t}$ between the plates equals the conduction current. Write the Ampere-Maxwell law.

Try it first, then show the solutionSolution

For a loop round the wire, a flat surface pierced by the wire gives $\oint B \cdot d l = μ_{0} i_{c}$ ; a balloon-shaped surface through the gap is crossed by no current and gives zero: a contradiction. (1)
Between plates of area A, $E = \frac{q}{ε _{0} A}$ and $Φ_{E} = E A = \frac{q}{ε _{0}}$ , so $ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i_{c}$ . (1)
Maxwell's displacement current $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ fills the gap: $\oint B \cdot d l = μ_{0} (i_{c} + i_{d})$ . (1)

Answer: The two surfaces give different results; $ε_{0} d Φ_{E} / d t = i_{c}$ ; Ampere-Maxwell law: ∮B·dl = μ₀(i_c + i_d).

Example 3: Rate of change of field · 2 marks · Very short answerHarder board level

The plates of a parallel plate capacitor have an area of 0.02 m². Find the rate at which the electric field between them must change to give a displacement current of 0.1 A.

Try it first, then show the solutionSolution

$i_{d} = ε_{0} A \frac{d E}{d t}$ . (1)
$\frac{d E}{d t} = \frac{0.1}{8.854 \times 1 0 ^{- 12} \times 0.02} \approx 5.65 \times 1 0^{11}$ V m⁻¹ s⁻¹. (1)

Answer: About $5.65 \times 1 0^{11}$ V m⁻¹ s⁻¹.

Example 4: Displacement current from changing voltage · 2 marks · Very short answerHarder board level

The voltage across a 2 μF capacitor rises steadily at 1000 V/s. Find the displacement current between its plates and show that it equals $C \frac{d V}{d t}$ .

Try it first, then show the solutionSolution

For plates of area A and separation d: $E = \frac{V}{d}$ , so $i_{d} = ε_{0} A \frac{d E}{d t} = \frac{ε _{0} A}{d} \frac{d V}{d t} = C \frac{d V}{d t}$ . (1)
$i_{d} = 2 \times 1 0^{- 6} \times 1000 = 2 \times 1 0^{- 3}$ A. (1)

Answer: 2 mA.

Example 5: Magnetic field between circular plates · 5 marks · Long answerHOTS / competency

A capacitor with circular plates of radius 10 cm is being charged by a current of 0.2 A.
(a) Find the displacement current between the plates. (1)
(b) Using the Ampere-Maxwell law, find the magnetic field at 5 cm from the axis, between the plates. (2)
(c) Find the field at 15 cm from the axis, and compare it with the field at the edge of the plates. (2)

Try it first, then show the solutionSolution

(a) $i_{d} = 0.2$ A (equal to the charging current). (1)
(b) A circular loop of radius r = 5 cm encloses the fraction $\frac{r ^{2}}{R ^{2}} = \frac{1}{4}$ of $i_{d}$ (assuming the field is uniform between the plates). (1)
$B \cdot 2 π r = μ_{0} \frac{i _{d} r ^{2}}{R ^{2}}$ gives $B = \frac{μ _{0} i _{d} r}{2 π R ^{2}} = \frac{4 π \times 1 0 ^{- 7} \times 0.2 \times 0.05}{2 π \times 0.01} = 2 \times 1 0^{- 7}$ T. (1)
(c) At r = 15 cm the loop encloses all of $i_{d}$ : $B = \frac{μ _{0} i _{d}}{2 π r} = \frac{2 \times 1 0 ^{- 7} \times 0.2}{0.15} \approx 2.67 \times 1 0^{- 7}$ T. (1)
At the edge (r = 10 cm), $B = 4 \times 1 0^{- 7}$ T, the largest value; B rises linearly up to the edge and then falls as $\frac{1}{r}$ . (1)

Answer: (a) 0.2 A; (b) $2 \times 1 0^{- 7}$ T; (c) about $2.67 \times 1 0^{- 7}$ T, less than the edge value of $4 \times 1 0^{- 7}$ T.

Example 6: Capacitor in an AC circuit · 4 marks · Case-basedHarder board level

Read the passage and answer.

A 10 μF capacitor is connected to a 220 V, 50 Hz supply.
(a) Find the rms current in the connecting wires. (1)
(b) What is the rms displacement current between the plates? (1)
(c) Is there a magnetic field in the gap between the plates? Explain. (1)
(d) What would the displacement current be with a steady 220 V DC supply, after the capacitor has charged? (1)

Try it first, then show the solutionSolution

(a) $I = V ω C = 220 \times 2 π \times 50 \times 1 0^{- 5} \approx 0.69$ A. (1)
(b) The same, about 0.69 A rms, since current is continuous. (1)
(c) Yes: the displacement current due to the changing electric field produces a magnetic field around it, just like a conduction current. (1)
(d) Zero: the field between the plates is constant. (1)

Answer: (a) about 0.69 A; (b) about 0.69 A; (c) yes, from the changing electric field; (d) zero.

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 in the gap.
The word 'current' misleads.
Do this: Say it is due to a changing electric field; no charge crosses the gap.
Usually costs: 1 mark.
Mistake: Saying displacement current exists in a fully charged capacitor on DC.
Forgetting it needs a changing field.
Do this: No change in E, no displacement current.
Usually costs: 1 mark.
Mistake: Writing i_d = ε₀ Φ_E (without the rate).
Dropping d/dt.
Do this: i_d = ε₀ dΦ_E/dt.
Usually costs: 1 mark.
Mistake: Explaining the inconsistency without the two surfaces.
Stating the conclusion only.
Do this: Describe the flat surface (crossed by current) and the balloon surface (through the gap).
Usually costs: 1 mark in the 3-mark answer.

Exam tips

Draw the charging capacitor with the loop, flat surface and balloon surface; it makes the 2-3 mark explanation clear.
End with the Ampere-Maxwell law and the statement that current is continuous.
Displacement current was set in 2023 (55/1/1 Q8 and Q20), 2024 (55/1/1 Q8), 2025 (55/7/1 Q25) and 2026 (55/5/1 Q25).
The syllabus asks for the basic idea; keep derivations to the capacitor case.
Link it to EM waves in one line when asked about its significance.

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
Displacement current arises from
(a) the flow of free electrons (b) a changing magnetic field (c) a changing electric field (d) a steady electric field
Hint
i_d = ε₀ dΦ_E/dt.
Answer
(c) a changing electric field.
Q2 · 1 mark · Very short answerBoard level
Write the expression for displacement current and its SI unit.
Hint
Electric flux.
Answer
$i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ ; unit ampere.
Q3 · 1 mark · Very short answerHarder board level
The electric flux through a surface changes at $1 0^{9}$ V m s⁻¹. Find the displacement current through it.
Hint
Multiply by ε₀.
Answer
$i_{d} = 8.854 \times 1 0^{- 12} \times 1 0^{9} \approx 8.85 \times 1 0^{- 3}$ A.
Q4 · 1 mark · Very short answerHarder board level
The voltage across a 5 pF capacitor changes at $1 0^{6}$ V/s. Find the displacement current.
Hint
i_d = C dV/dt.
Answer
$i_{d} = 5 \times 1 0^{- 12} \times 1 0^{6} = 5 \times 1 0^{- 6}$ A.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A magnetic field exists in the region between the plates of a capacitor while it is being charged.
Reason (R): The changing electric field between the plates gives rise to a displacement current, which produces a magnetic field.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Ampere-Maxwell law.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
Show that the total current (conduction plus displacement) is the same in the wires and between the plates of a charging capacitor.
Hint
Compare i_c in the wire with ε₀ dΦ_E/dt in the gap.
Answer
In the wire, i_c = dq/dt and i_d ≈ 0. In the gap, i_c = 0 and $i_{d} = ε_{0} \frac{d}{d t} (\frac{q}{ε _{0}}) = \frac{d q}{d t}$ . So the sum i_c + i_d is dq/dt in both places: current is continuous.
Q7 · 2 marks · Very short answerHOTS / competency
Explain why Maxwell's addition of displacement current was essential for predicting electromagnetic waves.
Hint
Think of Faraday's law alongside.
Answer
Faraday's law shows that a changing magnetic field creates an electric field. Displacement current means a changing electric field creates a magnetic field. With both, each changing field regenerates the other, so the disturbance can travel through empty space as an electromagnetic wave, with speed 1/√(μ₀ε₀) = c.

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.Expression for displacement current?Show answer
i_d = ε₀ dΦ_E/dt.
2.State the Ampere-Maxwell law.Show answer
∮B·dl = μ₀(i_c + i_d).
3.Displacement current between the plates of a capacitor charged by current I?Show answer
I.
4.Displacement current in a fully charged capacitor on DC?Show answer
Zero.
5.What does a changing electric field produce?Show answer
A magnetic field.
6.Who introduced displacement current?Show answer
Maxwell.

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. That is why it is called a current and appears alongside conduction current in the Ampere-Maxwell law.

Why was displacement current needed?

Without it, Ampere's law gives two different answers for the same loop around a charging capacitor, depending on which surface is chosen. Adding ε₀ dΦ_E/dt makes the law consistent and current continuous.

Where does displacement current exist?

Wherever an electric field is changing: between capacitor plates while charging or in AC circuits, and everywhere in an electromagnetic wave.

How is displacement current related to EM waves?

A changing electric field produces a magnetic field (displacement current), and a changing magnetic field produces an electric field (Faraday's law). Together they let the fields sustain each other as a travelling wave.

How much detail is needed for CBSE 2026-27?

The syllabus asks for the basic idea of displacement current: why it is needed, its expression i_d = ε₀ dΦ_E/dt and the Ampere-Maxwell law, usually illustrated with a charging capacitor.