Definition
Displacement current
Current associated with changing electric field: I_d = ε₀ dΦ_E/dt. Maxwell added this to Ampère's law to make it consistent with charge conservation.
-- NCERT Class 12 Physics, Ch. 8, p. 202Displacement Current
8 MCQs1 revision card9-step worked example
Source: NCERT Electromagnetic WavesPYQ coverage: NEET 2021, 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Displacement Current, explained for NEET
The distractor that costs marks here is treating displacement current as a flow of charge. It is not. Nothing crosses the gap between capacitor plates. The name is historical, and the exam exploits it.
Ampère's law in its original form fails for a charging capacitor. Take a loop around the wire and stretch the surface bounded by it. Pass the surface through the wire and you enclose conduction current I. Balloon the same surface so it passes between the plates instead, enclosing no charge flow at all, and the law gives zero. Same loop, same instant, two answers. Maxwell's fix, defined in NCERT Class 12 Physics Chapter 8 on page 202, is that a changing electric flux between the plates is itself a source of magnetic field:
I_d = ε₀ dΦ_E/dt
Between the plates the flux rises exactly fast enough that I_d equals the conduction current in the wire. Continuity is restored — the total current I_c + I_d is the same through every surface bounded by that loop.
Two consequences NEET tests directly. First, the magnetic field just outside a charging capacitor is what you would get if the wire ran straight through: the displacement current substitutes for the conduction current, with the same magnitude. Second, since I_d depends on dΦ_E/dt and not on Φ_E, a capacitor sitting at a constant voltage carries zero displacement current no matter how large the charge stored on it.
Sign and direction follow the changing field, not the wire. When the capacitor discharges, dΦ_E/dt reverses and so does I_d.
Watch out for the value of ε₀ in problems that hand you dΦ_E/dt in V·m/s. The product ε₀ × (V·m/s) is amperes directly; no unit gymnastics, but the order of magnitude (10⁻¹²) makes answers small, and options built at 10⁻⁹ and 10⁻¹⁵ sit either side of the right one.
Can you answer these Displacement Current MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Maxwell's modification of Ampère's circuital law, displacement current arises from
Show answer and why every option is right or wrong
Answer: C. C is correct. NCERT Class 12 Physics, Chapter 8, page 202 defines displacement current as the term arising from a time-varying electric flux, I_d = ε₀ dΦ_E/dt.
Why A is wrong: A is wrong because no charge crosses the gap at all; the region between the plates is empty of conduction. This is the conflation the pattern's chief distractor exploits.
Why B is wrong: B is wrong because displacement current exists between plates in vacuum, where there are no bound charges to polarise. A dielectric adds polarisation current, but is not the source of the displacement term.
Why D is wrong: D is wrong because a changing magnetic flux drives an induced EMF through Faraday's law, a separate equation. The displacement term is the electric-flux counterpart.
MCQ 2Concept TrapPractice
A parallel-plate capacitor is held at a steady potential difference of 12 V by a battery, with the circuit in steady state. The displacement current between its plates is
Show answer and why every option is right or wrong
Answer: A. A is correct. I_d = ε₀ dΦ_E/dt depends on the rate of change of flux, not on its value, so a constant flux gives zero displacement current, consistent with the definition on page 202 of NCERT Class 12 Physics, Chapter 8.
Why B is wrong: B is wrong because the presence of a dielectric is irrelevant to whether displacement current exists; the term is defined for vacuum in the first place.
Why C is wrong: C is wrong because in the steady state no current flows in this branch either — the capacitor is fully charged, so both the conduction and the displacement current are zero.
Why D is wrong: D is wrong because I_d responds to dΦ_E/dt, not to Φ_E. A large stored charge held constant produces no displacement current.
MCQ 3Direct ApplicationPractice
The electric flux between the plates of a capacitor changes at a uniform rate of 2.0 × 10⁶ V·m/s. Taking ε₀ = 8.85 × 10⁻¹² F/m, the displacement current is
Show answer and why every option is right or wrong
Answer: B. B is correct. I_d = ε₀ dΦ_E/dt = (8.85 × 10⁻¹²)(2.0 × 10⁶) = 1.77 × 10⁻⁵ A ≈ 1.8 × 10⁻⁵ A, applying the relation given on page 202 of NCERT Class 12 Physics, Chapter 8.
Why A is wrong: A is wrong because it misplaces the decimal exponent by two, most often by treating ε₀ as 8.85 × 10⁻¹⁰ F/m.
Why C is wrong: C is wrong because it halves the result, ½ε₀ dΦ_E/dt, borrowing the ½ from the energy density ½ε₀E²; the displacement current has no ½.
Why D is wrong: D is wrong because it drops the 10⁶ of the flux rate, multiplying ε₀ by 2.0 alone.
MCQ 4CalculationPractice
A parallel-plate capacitor of capacitance 5.0 µF is being charged so that the potential difference across it increases at a steady rate of 4.0 V/s. The displacement current between its plates is
Show answer and why every option is right or wrong
Answer: C. C is correct. Between the plates ε₀ dΦ_E/dt equals the conduction current in the leads, and for a capacitor I = C dV/dt = (5.0 × 10⁻⁶)(4.0) = 2.0 × 10⁻⁵ A. This equality is the content of Maxwell's correction as set out on page 202 of NCERT Class 12 Physics, Chapter 8.
Why A is wrong: A is wrong because it multiplies ε₀ by dV/dt directly, (8.85 × 10⁻¹²)(4.0) = 3.5 × 10⁻¹¹ A, ignoring that the flux rate is not the voltage rate — the plate area and separation enter through C.
Why B is wrong: B is wrong because it divides C by dV/dt instead of multiplying: 5.0 × 10⁻⁶/4.0 = 1.25 × 10⁻⁶.
Why D is wrong: D is wrong because it reads 5.0 µF as 5.0 pF, taking C = 5.0 × 10⁻¹² F: (5.0 × 10⁻¹²)(4.0) = 2.0 × 10⁻¹¹ A.
MCQ 5Concept TrapPractice
An Ampèrian loop of fixed radius encircles the wire feeding a charging parallel-plate capacitor. Surface S₁ bounded by this loop is pierced by the wire; surface S₂ bounded by the same loop bulges out and passes through the gap between the plates. Which statement is correct?
Show answer and why every option is right or wrong
Answer: B. B is correct. The whole purpose of the displacement term is consistency: I_c through S₁ equals I_d through S₂, so the modified law gives one magnetic field for the loop regardless of the surface chosen — the argument made on page 202 of NCERT Class 12 Physics, Chapter 8.
Why A is wrong: A is wrong because it is precisely the inconsistency Maxwell removed. S₂ encloses no conduction current but does enclose a changing electric flux, which contributes I_d.
Why C is wrong: C is wrong because ε₀ is not a ratio between two fields here; I_d is constructed to equal I_c in magnitude, so the two surfaces give equal fields, not fields differing by a factor.
Why D is wrong: D is wrong because the modified law applies to time-varying currents; that extension is the point of the displacement term rather than a limitation on it.
MCQ 6Easy RecallPractice
The SI unit of the quantity ε₀ dΦ_E/dt is
Show answer and why every option is right or wrong
Answer: D. D is correct. The quantity is a current — displacement current — and carries the ampere, as its definition on page 202 of NCERT Class 12 Physics, Chapter 8 requires for the term to sit alongside conduction current in Ampère's law.
Why A is wrong: A is wrong because the volt measures potential difference. Electric flux carries V·m, and its time derivative scaled by ε₀ reduces to the ampere, not the volt.
Why B is wrong: B is wrong because the weber is the unit of magnetic flux, so weber per second is the unit of induced EMF in Faraday's law — a different equation.
Why C is wrong: C is wrong because the tesla measures magnetic flux density, which is what the modified Ampère's law produces from this term, not what the term itself is measured in.
MCQ 7CalculationPractice
A parallel-plate capacitor with circular plates of radius R is charging. At a point inside the gap at perpendicular distance r from the axis, with r < R, the magnetic field is proportional to
Show answer and why every option is right or wrong
Answer: B. B is correct. The displacement current is spread uniformly over the plate area, so a loop of radius r encloses I_d × (r²/R²); dividing by the circumference 2πr leaves a field proportional to r — the same enclosed-fraction argument NCERT Class 12 Physics, Chapter 8 (page 202) applies once I_d is treated as a current.
Why A is wrong: A is wrong because 1/r holds outside the plate radius, where the full displacement current is enclosed. Inside the gap only part of it is.
Why C is wrong: C is wrong because it stops at the enclosed current, which does scale as r², and forgets to divide by the loop circumference 2πr.
Why D is wrong: D is wrong because the absence of charge flow does not mean the absence of a magnetic field; the changing electric flux sources one, which is exactly Maxwell's point.
MCQ 8Direct ApplicationPractice
A capacitor initially charging is switched to discharge through the same circuit, with the magnitude of the rate of change of electric flux unchanged. Compared with the charging phase, the displacement current between the plates now
Show answer and why every option is right or wrong
Answer: A. A is correct. I_d = ε₀ dΦ_E/dt takes its sign from the derivative, so reversing the sense in which the flux changes reverses I_d while leaving its magnitude set by the unchanged rate — direct from the definition on page 202 of NCERT Class 12 Physics, Chapter 8.
Why B is wrong: B is wrong because the direction is tied to the sign of dΦ_E/dt, which flips when the field between the plates starts decreasing instead of increasing.
Why C is wrong: C is wrong because ε₀ is already contained in the value of I_d; it does not act a second time as a scaling factor when the process reverses.
Why D is wrong: D is wrong because a discharging capacitor has a falling electric flux, which is a non-zero rate of change and therefore a non-zero displacement current.
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Displacement Current: quick recall before you leave
How do you solve a Displacement Current question? A worked example
Pattern: NEET pattern: displacement current problem — capacitor with changing voltage; displacement current equals the capacitor's conduction current. Observed 2021 and 2025; skill multi_step; ~75 s.
- 1
Given
Parallel-plate capacitor, circular plates of radius R = 6.0 × 10⁻² m, separation d = 2.0 × 10⁻³ m. Potential difference across the plates rises uniformly at dV/dt = 5.0 × 10² V/s. ε₀ = 8.85 × 10⁻¹² F/m. Vacuum between the plates.
- 2
Required
The displacement current between the plates.
- 3
Concept
No charge crosses the gap, yet the magnetic field around the leads does not stop at the plates. The changing electric flux between them acts as a current for the purposes of Ampère's law. That is the displacement current defined in NCERT Class 12 Physics, Chapter 8, page 202.
- 4
Formula
I_d = ε₀ dΦ_E/dt, with Φ_E = E·A and E = V/d for a parallel-plate geometry, giving
I_d = ε₀ (A/d) dV/dt, where A = πR². - 5
Substitution
A = π(6.0 × 10⁻²)² = π × 3.6 × 10⁻³ m²
I_d = (8.85 × 10⁻¹²) × (π × 3.6 × 10⁻³ / 2.0 × 10⁻³) × (5.0 × 10²) - 6
Calculation
A = 1.131 × 10⁻² m²
A/d = 1.131 × 10⁻² / 2.0 × 10⁻³ = 5.655 m
ε₀ (A/d) = (8.85 × 10⁻¹²)(5.655) = 5.005 × 10⁻¹¹ F
I_d = (5.005 × 10⁻¹¹)(5.0 × 10²) = 2.50 × 10⁻⁸ A
Note on exact quantities: π is a mathematical constant and the 2 in πR² is a counting exponent; neither contributes to the significant-figure count. The measured inputs (6.0 × 10⁻², 2.0 × 10⁻³, 5.0 × 10²) each carry two significant figures, and ε₀ is quoted to three, so the answer is reported to two. - 7
Final answer
I_d = 2.5 × 10⁻⁸ A.
Note that ε₀ (A/d) is just the capacitance C = 5.0 × 10⁻¹¹ F, so the same number falls out of I = C dV/dt. The displacement current between the plates equals the conduction current in the wire, which is the whole point. - 8
Common trap
The pattern's recorded distractor is treating displacement current as a real flow of charge — students look for electrons crossing the 2.0 mm gap, find none, and answer zero. The second failure mode is arithmetic: dropping the factor A/d and computing ε₀ dV/dt alone, which returns 4.4 × 10⁻⁹ A and is offered as an option. The geometry enters through the flux; voltage rate alone is not flux rate.
- 9
Similar NEET-style question
A parallel-plate capacitor with square plates of side 4.0 × 10⁻² m and separation 1.0 × 10⁻³ m is connected to a source whose voltage falls uniformly at 2.0 × 10³ V/s. Find the magnitude and sense of the displacement current between the plates. *(Answer: 2.8 × 10⁻⁸ A, directed opposite to the charging-phase displacement current, since dΦ_E/dt is negative.)*
What to remember before solving Displacement Current questions
Which Displacement Current formulas do you need for NEET?
1 formula — click to collapse
Displacement current
Current associated with changing electric flux. Closes Ampere's law for time-varying fields.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_d | displacement current | A |
| Phi_E | electric flux | V*m |
| eps0 | permittivity | F/m |
Valid when
- Time-varying E field
- Capacitor or vacuum region
More in Electromagnetic Waves: 2 exam traps and mistakes · 3 formulas · 2 question patterns from its other lessons.
Displacement Current questions from past NEET papers
2 questions from NEET 2021, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1Zero between the plates and non-zero outside
2Zero at all places
3Constant between the plates and zero outside the plates
4Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
NTA Answer: Option 4(final)
NEET 2021
1I_d = V₀ωC sinωt
2I_d = V₀ωC cosωt
3I_d = (V₀/(ωC)) cosωt
4I_d = (V₀/(ωC)) sinωt
NTA Answer: Option 2(final)
How does NEET ask about Displacement Current?
1 recurring pattern from past papers — click to collapse
Capacitor with changing voltage: displacement current I_d = ε_0 dΦ_E/dt = capacitor I.
Multi StepMedium
Common distractors
treats displacement as real current
Conflates conduction with displacement
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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