Displacement current

Pick up any Class 11 textbook and Ampere's circuital law looks bulletproof: draw a loop, add up the magnetic field going around it, and that circulation tells you exactly how much current pierces any surface stretched across the loop. It never seems to matter which surface you choose — a flat one, a bulging one, a weirdly dented one — the answer to "how much current threads it" was always supposed to come out the same. Then Maxwell looked at something as ordinary as a capacitor being charged on a circuit board, and the law broke. Draw a loop around the wire feeding a charging capacitor. Cap it with a flat disc that the wire pokes through, and Ampere's law happily reports the conduction current in the wire. Now cap the very same loop with a surface that bulges forward and slides between the capacitor's plates instead of cutting the wire — and suddenly no charge crosses that surface at all, because nothing physically jumps the gap between the plates. Same loop, same law, two different surfaces, two wildly different answers: current, and zero. Something in the "beautiful" law was quietly incomplete, and nobody had noticed for decades because nobody had bothered to check a capacitor. Maxwell's fix is one of the most elegant patches in all of physics: he noticed that even though no charge crosses the gap, the electric field between the plates is changing every instant the capacitor charges — and he proposed that a changing electric field should count as a current too, for exactly the purpose of feeding Ampere's law. That invented quantity is the displacement current, and it is the missing piece that not only rescues Ampere's law but also opens the door to electromagnetic waves themselves.

Key takeaways

Ampere's original circuital law gives inconsistent answers for a capacitor-charging circuit depending on which surface bounded by the loop you choose to use.
The inconsistency shows up specifically between the plates of the capacitor, where conduction current is zero but the electric field is changing with time.
Maxwell proposed a displacement current $I_d = \varepsilon_0\dfrac{d\phi_E}{dt}$ , sourced by a changing electric flux rather than by moving charge, to patch the gap.
Displacement current is not a flow of charge; it is real in its physical effect — a changing electric field genuinely produces a magnetic field, exactly like a conduction current would.
At every instant of charging, the displacement current between the plates numerically equals the conduction current in the connecting wire: $I_d = I_c$ .
With this addition, Ampere's law becomes the Ampere–Maxwell law, and it gives the same, consistent answer no matter which surface bounded by the loop is used.
Key idea: because a time-varying electric flux between the capacitor's plates produces a magnetic field just as a real current would, Maxwell added a displacement current $I_d=\varepsilon_0\,d\phi_E/dt$ (numerically equal to the conduction current at every instant) to Ampere's law, making it self-consistent and setting up the E–B symmetry that electromagnetic waves are built on.

The core insight

A changing electric flux through any region acts as a source of magnetic field exactly like a conduction current does, so Maxwell's displacement current $I_d=\varepsilon_0\,d\phi_E/dt$ — always numerically equal to the conduction current feeding a charging capacitor — completes Ampere's law and makes it consistent for every possible surface.

Where it fits in the syllabus

Displacement current is the opening idea of the CBSE Class 12 chapter Electromagnetic Waves, and the 2025-26 prescribed scope keeps it strictly qualitative: you are expected to know why the original Ampere's circuital law is incomplete for a capacitor-charging circuit, what displacement current physically represents, and that it completes Ampere's law — but not to reproduce Maxwell's four equations or a formal calculus derivation of $I_d=\varepsilon_0\,d\phi_E/dt$ from first principles. This section keeps that boundary: the formula is used and its plausibility is shown through the ordinary parallel-plate capacitor (algebra only, no vector calculus), never presented as a memorised proof you must reproduce verbatim in the exam. Everything that follows — the speed of light $c=1/\sqrt{\mu_0\varepsilon_0}$ , the transverse E–B wave picture, and the electromagnetic spectrum — depends on this one idea: fields can source fields, without any charge in sight.

The physical picture

Start with the ordinary Ampere's circuital law from Moving Charges and Magnetism, which relates the circulation of $\vec B$ around a closed loop to the current threading a surface bounded by that loop.

\oint \vec B \cdot d\vec l = \mu_0 I

where: $\oint \vec B\cdot d\vec l$ is the circulation of the magnetic field around a closed loop, $\mu_0$ is the permeability of free space, and $I$ is the current through any surface bounded by the loop. Physically this says a current always has a magnetic field curling around it — completely reliable for a straight current-carrying wire, where any surface you cap the loop with (flat, tilted, dented) gets pierced by the same wire and reports the same $I$ . Now put a capacitor in that wire and charge it. Choose the bulging surface that threads between the plates instead of across the wire. No charge crosses this surface — the plates never touch — so a literal reading of the law says the enclosed current is zero, and yet the same loop, capped with the flat surface through the wire, insists the current is $I$ . Maxwell's resolution begins by naming what is happening in the gap: the electric field $E$ between the plates is climbing every instant the capacitor charges, which means the electric flux through that bulging surface, $\phi_E$ , is also climbing. He proposed that this changing flux itself acts as a source term for $\vec B$ , exactly as if it were a current — the displacement current:

I_d = \varepsilon_0\,\frac{d\phi_E}{dt}

where: $\varepsilon_0$ is the permittivity of free space, $\phi_E=\int \vec E\cdot d\vec A$ is the electric flux through the chosen surface, and $I_d$ is measured in amperes just like ordinary current. Physically, $I_d$ is not moving charge at all — it is a bookkeeping current, born purely of a time-varying $E$ -field, that nonetheless produces a real, measurable magnetic field. Adding it to the conduction current gives the corrected Ampere–Maxwell law,

\oint \vec B \cdot d\vec l = \mu_0\left(I_c + I_d\right)

so that now, for the flat surface through the wire, $I_d=0$ and only $I_c$ contributes; for the bulging surface between the plates, $I_c=0$ and only $I_d$ contributes — and because $I_d=I_c$ at every instant (shown below), both surfaces finally agree. This is the specific "fields can source fields" symmetry — a changing $E$ makes a $B$ , just as Faraday's law says a changing $B$ makes an $E$ — that later lets the two fields regenerate each other and travel as a self-sustaining wave at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ .

Figure

A charging capacitor with a wire loop around it, one flat surface cutting the wire carrying conduction current, and a second bulging surface passing between the plates carrying only displacement current, with both surfaces bounded by the same loop and giving the same enclosed current

Derivation AWhy displacement current equals conduction current for a charging capacitor

The goal here is modest and entirely algebraic: show, for the one concrete case CBSE actually cares about — a parallel-plate capacitor being charged — that the displacement current $I_d=\varepsilon_0\,d\phi_E/dt$ works out to be numerically identical to the conduction current $I_c=dQ/dt$ flowing in the wire, so the "two surfaces must agree" claim used above is not just asserted but checked. This uses only the elementary parallel-plate field result $E=\sigma/\varepsilon_0=Q/(\varepsilon_0 A)$ from Electrostatic Potential and Capacitance, plus one rate of change — no vector calculus and none of Maxwell's four equations, exactly the ceiling this syllabus allows.

Step 1. Write the charge and field on the plates. For a parallel-plate capacitor of plate area $A$ carrying instantaneous charge $Q$ , the field between the plates is uniform and given by $E=\dfrac{Q}{\varepsilon_0 A}$ , so the electric flux through a surface between the plates (area $A$ , field perpendicular to it) is $\phi_E = EA = \dfrac{Q}{\varepsilon_0}$ .

Step 2. Differentiate the flux with respect to time. Since $A$ and $\varepsilon_0$ are constants of the setup while $Q$ grows as the capacitor charges, $\dfrac{d\phi_E}{dt} = \dfrac{1}{\varepsilon_0}\dfrac{dQ}{dt}$ .

Step 3. Substitute into Maxwell's definition of displacement current. $I_d = \varepsilon_0\dfrac{d\phi_E}{dt} = \varepsilon_0\cdot\dfrac{1}{\varepsilon_0}\dfrac{dQ}{dt} = \dfrac{dQ}{dt}$ .

Step 4. Identify the right-hand side. But $\dfrac{dQ}{dt}$ is exactly the conduction current $I_c$ flowing in the wire feeding the plates — the rate at which charge is delivered to the capacitor. So $I_d = I_c$ at every instant of charging, confirming that the two surfaces of the Ampere loop really must give the same answer, and that displacement current is not a free-floating idea but pinned down by ordinary capacitor algebra you already know. Note: CBSE does not require reproducing this four-line algebra from memory — the syllabus wants only the qualitative statement $I_d=I_c$ and why it matters — but seeing where the equality comes from removes the "why should I just believe that" feeling.

Worked examples

Board level - conduction current tells you the displacement current

A parallel-plate capacitor with circular plates of area $A = 500\ \text{cm}^2 = 0.05\ \text{m}^2$ is being charged, and at a certain instant the conduction current in the connecting wire is measured as $I = 2\ \text{A}$ . Find the displacement current between the plates at that instant, and the rate at which the electric field between the plates is changing.

Step 1. State the consistency requirement. Because the Ampere–Maxwell law must give the same enclosed current whichever surface bounded by the loop is chosen, the displacement current between the plates equals the conduction current in the wire at every instant: $I_d = I_c$ .

Step 2. Read off $I_d$ . Given $I_c = 2\ \text{A}$ , immediately $I_d = 2\ \text{A}$ . Even though no charge crosses the gap between the plates, the changing electric field there behaves, for the purposes of Ampere's law, exactly as if a current of 2 A were flowing across the gap.

Step 3. Use $I_d=\varepsilon_0 A\,\dfrac{dE}{dt}$ to find the rate of change of the field. Since $\phi_E = EA$ for the uniform field between the plates, $I_d = \varepsilon_0 A\,\dfrac{dE}{dt}$ , so $\dfrac{dE}{dt} = \dfrac{I_d}{\varepsilon_0 A} = \dfrac{2}{(8.85\times10^{-12})(0.05)}$ .

Step 4. Compute and state the answer. $\dfrac{dE}{dt} = \dfrac{2}{4.425\times10^{-13}} \approx 4.52\times10^{12}\ \text{V/(m}\,\cdot\,\text{s)}$ . The displacement current is $I_d = 2\ \text{A}$ , and the electric field between the plates is increasing at about $4.52\times10^{12}\ \text{V/(m}\,\cdot\,\text{s)}$ — a reminder of how fast the field must climb for a laboratory-scale current to appear from such a tiny constant as $\varepsilon_0$ .

Competitive level - magnetic field between the plates from displacement current

The same capacitor now has circular plates of radius $R = 6\ \text{cm} = 0.06\ \text{m}$ , still charging with $I_c = I_d = 2\ \text{A}$ (same numbers as the board-level example). Find (a) the displacement current density $J_d$ between the plates, and (b) the magnitude of the magnetic field at a point midway between the plates, at a radial distance $r = 3\ \text{cm}$ from the central axis. (This uses the quoted, symmetry-based Ampere–Maxwell result $B = \dfrac{\mu_0 I_d\, r}{2\pi R^2}$ for $r \le R$ — a standard competitive-exam extension one level above the CBSE board ceiling, given here without re-derivation.)

Step 1. Find the displacement current density. Assuming $I_d$ is spread uniformly over the circular plate area, $J_d = \dfrac{I_d}{\pi R^2} = \dfrac{2}{\pi (0.06)^2} = \dfrac{2}{0.01131} \approx 176.8\ \text{A/m}^2$ .

Step 2. Quote the magnetic field formula for a point inside the plate radius. For $r \le R$ , symmetry (the same cylindrical symmetry used for a straight current-carrying wire, applied to the disc of displacement current) gives $B = \dfrac{\mu_0 I_d\, r}{2\pi R^2}$ ; using $\mu_0/2\pi = 2\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ , this simplifies to $B = (2\times10^{-7})\dfrac{I_d\, r}{R^2}$ .

Step 3. Substitute the numbers. $B = (2\times10^{-7})\dfrac{(2)(0.03)}{(0.06)^2} = (2\times10^{-7})\dfrac{0.06}{0.0036} = (2\times10^{-7})(16.67) \approx 3.33\times10^{-6}\ \text{T}$ .

Step 4. State the answers. $J_d \approx 176.8\ \text{A/m}^2$ and $B \approx 3.33\ \mu\text{T}$ at $r=3\ \text{cm}$ — a tiny but real and, in Maxwell's era, eventually measurable magnetic field, generated with no charge whatsoever crossing the gap. Timing note: in a competitive exam, recognise instantly that $I_d=I_c$ is given (or trivially equal) before touching the $B$ -field formula — don't waste time re-deriving that equality from flux integrals under time pressure; quote it, then plug into the symmetry result for $B$ and move on.

Common traps

Thinking displacement current involves actual moving charge between the capacitor plates — it does not; it is defined purely through the rate of change of electric flux, and no charge crosses the gap at any point.
Assuming displacement current is an exotic effect confined to capacitors — it exists wherever the electric flux through any region changes with time, capacitor or not, even in a plain patch of vacuum.
Trying to reproduce a full calculus derivation of Maxwell's four equations in an exam answer — the CBSE syllabus wants only the qualitative idea and the "two-surfaces-must-agree" consistency argument, not the formal proof.
Confusing displacement current with eddy currents or induced currents from Electromagnetic Induction — those involve real moving charge in a conductor; displacement current involves no conductor and no real charge flow at all.
Forgetting that $I_d=I_c$ holds only instant-by-instant during charging (or discharging) — once the capacitor is fully charged, both currents drop to zero together, not just one of them.

You should be able to…

Explain why applying the original Ampere's circuital law to a capacitor-charging circuit gives inconsistent results depending on the chosen surface.
State what displacement current physically represents: a changing electric flux acting as a source of magnetic field, with no real charge motion involved.
Show, using $\phi_E=Q/\varepsilon_0$ for a parallel-plate capacitor, why $I_d$ works out numerically equal to the conduction current $I_c$ feeding it.
Use $I_d=\varepsilon_0\,d\phi_E/dt$ (or $I_d=\varepsilon_0 A\,dE/dt$ ) to compute displacement current or rate of change of field in a numerical problem.
Distinguish displacement current from ordinary conduction current and from induced (eddy) current.
Explain why displacement current is the essential ingredient that lets electromagnetic waves exist later in the chapter.

Ask the tutor: if the "two different surfaces of the same loop must agree" argument still feels shaky, ask the AI tutor to walk through the flat-surface-versus-bulging-surface picture step by step with the capacitor diagram in front of you.

If no charge crosses the gap between the plates, how can there be a current there at all?

Because "current" in Ampere's law is really being used as shorthand for "whatever produces circulating magnetic field," and Maxwell discovered that a changing electric flux does exactly that job, even with zero moving charge. He kept the name "current" and the ampere units because the mathematics and the physical effect (a real magnetic field appearing around the gap) are identical to those of ordinary current — but you should picture it as a changing field, never as charge sneaking across.

Do I need to memorise the derivation of I_d = epsilon-zero d(phi-E)/dt for my board exam?

No. The CBSE 2025-26 syllabus explicitly keeps the formal derivation of Maxwell's equations and the displacement current formula out of the required theory — you are expected to state what displacement current represents and quote the consistency result $I_d=I_c$ , not derive it from scratch. The capacitor-algebra check in this article is there to build intuition, not to be reproduced word for word in an answer.

Why does this topic matter if it is "just" a fix to a law I already learned?

Because that one fix is the seed of the entire rest of the chapter. Once a changing electric field is allowed to produce a magnetic field (displacement current), and a changing magnetic field is already known to produce an electric field (Faraday's law), the two fields can regenerate each other continuously and travel through empty space as a self-propagating electromagnetic wave — light, radio, X-rays, all of it. Without displacement current, Maxwell's equations would predict no such wave at all.

Is displacement current only relevant while the capacitor is charging, or does it show up elsewhere?

It shows up anywhere the electric flux through a region changes with time — charging or discharging a capacitor is just the easiest example to picture. In an electromagnetic wave travelling through vacuum, for instance, the electric field at every point is constantly oscillating, so a displacement current (and its magnetic-field-producing effect) is present throughout the wave, not just near capacitor plates.