The Displacement Current

The Charging-Capacitor Paradox

Take a parallel-plate capacitor being charged by a time-dependent current $i(t)$, and ask for the magnetic field at a point P just outside it, near the connecting wire. Apply Ampere's circuital law, $\oint\vec{B}\cdot d\vec{l} = \mu_0 i(t)$, to a circular loop of radius r around the wire:

• Surface 1 — a flat disc spanning the loop, pierced by the wire: $B(2\pi r) = \mu_0 i(t)$. Fine.
• Surface 2 — a pot-shaped surface with the same circular mouth, whose bottom passes between the capacitor plates, touching no wire: enclosed current zero, so the law says $B(2\pi r) = 0$.
• A tiffin-box-shaped surface does the same.

Contradiction! The same point P has a field computed one way and zero computed another. Since Ampere's law produced it, Ampere's law must be incomplete — something must be added so that every surface with the same boundary gives the same answer.

Maxwell's Fix: The Displacement Current

Look at what does cross the gap-spanning surface: the electric field! Between plates of area A carrying charge Q, $E = \frac{Q}{A\varepsilon_0}$ (perpendicular to the surface), so the electric flux is

$$\Phi_E = EA = \frac{Q}{\varepsilon_0}$$

As the capacitor charges, Q changes, and

$$\varepsilon_0\frac{d\Phi_E}{dt} = \frac{dQ}{dt} = i$$

— exactly the missing current! Maxwell's resolution: a changing electric flux acts as a current, the displacement current

$$\boxed{i_d = \varepsilon_0\frac{d\Phi_E}{dt}}$$

The total current is $i = i_c + i_d$ (conduction + displacement), and the Ampere-Maxwell law reads

$$\boxed{\oint\vec{B}\cdot d\vec{l} = \mu_0 i_c + \mu_0\varepsilon_0\frac{d\Phi_E}{dt}}$$

Bookkeeping for the capacitor: outside (in the wire) $i_c = i$, $i_d = 0$; inside the gap $i_c = 0$, $i_d = i$. Every surface now reports the same total current — paradox dissolved.

Key Point: $i_d$ has all the physical effects of a real current — it produces magnetic fields. The field at a point M between the plates is measurably the same as at P just outside.

Consequences: A More Symmetric Electromagnetism

With the displacement current in place:

• Faraday's law: a time-varying magnetic field gives rise to an electric field.
• Ampere-Maxwell law: a time-varying electric field gives rise to a magnetic field.

Time-dependent electric and magnetic fields give rise to each other — the seed of electromagnetic waves (Section 3 onwards).

NCERT's careful footnote: the symmetry is still not perfect — there are no known magnetic monopoles (no magnetic analogue of charge as a source), which is why Gauss's law for magnetism keeps its zero.

[JEE Tip] For a parallel-plate capacitor, $i_d = \varepsilon_0 A\frac{dE}{dt} = C\frac{dV}{dt}$ — three interchangeable forms ($\Phi_E$, E, or V version). Choose whichever the data offers.

[NEET Important] In a region with steady fields (e.g. a steady current in a wire), $i_d = 0$: displacement current exists only while the electric field is changing. And in general both $i_c$ and $i_d$ can coexist in the same region — no medium is perfectly conducting or perfectly insulating.

Solved Examples

Example 1: Displacement current equals conduction current [NEET Numerical]

A capacitor is charged by a steady conduction current of 0.15 A in its leads. What is the displacement current between its plates?

Solution:

1. Between the plates, the changing Q gives $i_d = \varepsilon_0\,d\Phi_E/dt = dQ/dt$.
2. $dQ/dt$ is precisely the charging current: $i_d = 0.15$ A.
3. Continuity restored: the 'current' is 0.15 A everywhere around the circuit — conduction in the wires, displacement in the gap.

Example 2: From dV/dt [JEE Numerical]

A 1.0 $\mu$F parallel-plate capacitor has its potential difference rising at $5 \times 10^5$ V/s. Find the displacement current.

Solution:

1. $i_d = C\frac{dV}{dt}$ (capacitor form of $\varepsilon_0\,d\Phi_E/dt$).
2. $i_d = 10^{-6} \times 5 \times 10^5 = 0.5$ A.
3. Half an ampere flows 'through' the gap — as a changing field, not as charge.

Example 3: From dE/dt [JEE Numerical]

The electric field between circular plates of area $0.01$ m$^2$ changes at $10^{12}$ V m$^{-1}$ s$^{-1}$. Find $i_d$.

Solution:

1. $i_d = \varepsilon_0 A\frac{dE}{dt} = 8.85 \times 10^{-12} \times 0.01 \times 10^{12}$.
2. $i_d = 8.85 \times 10^{-2}$ A $\approx 88.5$ mA.
3. Enormous dE/dt values are needed for modest currents — $\varepsilon_0$ is tiny.

Example 4: The two-surface check

State the paradox the displacement current resolves, in two sentences.

Solution:

1. For the same Amperian loop near a charging capacitor, a flat surface (pierced by the wire) gives $B(2\pi r) = \mu_0 i$, while a pot-shaped surface through the gap gives zero — Ampere's law contradicts itself.
2. Adding $i_d = \varepsilon_0\,d\Phi_E/dt$ (which lives exactly where the wire doesn't) makes the total current through every spanning surface identical.

Example 5: B between the plates [JEE Numerical]

Circular capacitor plates of radius R = 6.0 cm carry a displacement current of amplitude 6.9 $\mu$A (uniformly distributed). Find the amplitude of B at r = 3.0 cm from the axis.

Solution:

1. Inside the gap, Ampere-Maxwell on a circle of radius r encloses the fraction $r^2/R^2$ of $i_d$: $B(2\pi r) = \mu_0 i_d\frac{r^2}{R^2}$.
2. $B = \frac{\mu_0 i_d r}{2\pi R^2} = \frac{4\pi \times 10^{-7} \times 6.9 \times 10^{-6} \times 0.03}{2\pi \times (0.06)^2}$.
3. $B = \frac{2 \times 10^{-7} \times 6.9 \times 10^{-6} \times 0.03}{3.6 \times 10^{-3}} \approx 1.15 \times 10^{-11}$ T.
4. Tiny but real — and measurable, confirming the displacement current's physical reality.

Example 6: An oscillating capacitor [JEE Numerical]

A 100 pF capacitor is connected to a source $V = 230\sin(300\,t)$ V. Find the peak displacement current.

Solution:

1. $i_d = C\frac{dV}{dt} = C \times 230 \times 300\cos(300t)$.
2. Peak: $i_d^{max} = 100 \times 10^{-12} \times 230 \times 300 = 6.9 \times 10^{-6}$ A.
3. The 6.9 $\mu$A of Example 5 — these two examples are one circuit.

Example 7: Where is each current?

For the charging capacitor, fill in: in the connecting wires, $i_c = ?$, $i_d = ?$; between the plates, $i_c = ?$, $i_d = ?$.

Solution:

1. Wires: $i_c = i$ (real charge flow), $i_d = 0$ (steady E inside a good conductor changes negligibly).
2. Gap: $i_c = 0$ (no charge crosses), $i_d = i$ (the changing E carries the baton).
3. Their sum $i_c + i_d = i$ is the same through every cross-section — current continuity, rescued.

Example 8: When is $i_d$ zero?

Give a situation with a non-zero electric field but zero displacement current, and one with $i_d \ne 0$ in empty space.

Solution:

1. Zero $i_d$: a steady electric field (e.g. inside a wire carrying steady DC, or a fully charged capacitor) — E exists but $dE/dt = 0$.
2. Non-zero $i_d$ in vacuum: any region where E varies with time — e.g. between charging plates, or in a passing electromagnetic wave — produces a magnetic field with no conduction current anywhere nearby.
3. That second case is precisely how EM waves sustain themselves.

Example 9: The symmetry argument

In what sense does the displacement current make electromagnetism 'more symmetrical' — and why not perfectly so?

Solution:

1. Faraday: changing B $\to$ E. Ampere-Maxwell: changing E $\to$ B. Each field now sources the other — a beautiful reciprocity.
2. The symmetry is imperfect because magnetic monopoles do not exist: electric fields have point sources (charges); magnetic fields do not.
3. Hence Gauss's law for E has $Q/\varepsilon_0$ on the right; Gauss's law for B has zero.

Example 10: Reading the Ampere-Maxwell law

Identify each term of $\oint\vec{B}\cdot d\vec{l} = \mu_0 i_c + \mu_0\varepsilon_0\,d\Phi_E/dt$ and when each dominates.

Solution:

1. Left side: circulation of B around a closed loop.
2. $\mu_0 i_c$: conduction-current source — dominates in wires and circuits.
3. $\mu_0\varepsilon_0\,d\Phi_E/dt$: displacement-current source — the only term in charging-capacitor gaps and in electromagnetic waves; both can coexist in ordinary (imperfect) media.