Formula of Displacement Current: I = ε₀ dΦ/dt with Examples

The formula of displacement current is $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ : multiply the permittivity of free space by how fast the electric flux through a surface is changing. Between the flat plates of a capacitor, where $\Phi_E = EA$ , it becomes $I_d = \varepsilon_0 A \frac{dE}{dt}$ .

No charge actually crosses the gap of a capacitor. Yet while the capacitor charges, a magnetic field curls around the gap exactly as it would around a wire. Displacement current is the term that accounts for this. It turns Ampere's circuital law into a rule that holds for changing fields too, and it is the step that predicts electromagnetic waves in Class 12. It belongs with the other electricity and magnetism formulas on this site.

Quick facts

Formula
$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$
Parallel plates
$I_d = \varepsilon_0 A \frac{dE}{dt} = C\frac{dV}{dt}$
SI unit
ampere (A)
Dimensional formula
[M⁰ L⁰ T⁰ I¹]
Type
Scalar, like conduction current
Corrected circuital law
$\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d)$

Reading the formula of displacement current

Symbols in the displacement current relation
Symbol	Meaning	SI unit
$I_d$	Displacement current	A
$\varepsilon_0$	Vacuum permittivity, close to 8.854 × 10⁻¹² F/m	F/m, same as C²/(N m²)
$\Phi_E$	Electric flux, $EA\cos\theta$	N m²/C
$\frac{d\Phi_E}{dt}$	Rate at which that flux changes	N m²/(C s)

A steady field gives zero displacement current, however strong the field is. Only change matters. In a medium other than vacuum, replace $\varepsilon_0$ by $\varepsilon = \varepsilon_r \varepsilon_0$ .

Why it equals the charging current

Take a parallel plate capacitor with plate area A and gap d. The gap field works out as $E = \frac{q}{\varepsilon_0 A}$ , so the flux through a surface in the gap is $\Phi_E = EA = \frac{q}{\varepsilon_0}$ . Differentiate with respect to time:

$\varepsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt} = I_c$

The displacement current in the gap equals the conduction current in the connecting wires. Current is continuous all round the circuit: charges carry it in the wires and the changing field carries it across the gap. Using $C = \frac{\varepsilon_0 A}{d}$ and $E = V/d$ , the same result reads $I_d = C\frac{dV}{dt}$ . Capacitance itself is covered on the capacitance formula page.

Worked problems

Example 1: field changing between plates

The plates of a capacitor have area 0.030 m². The field between them grows at 4.0 × 10¹² V/(m s). Find the displacement current.

$I_d = \varepsilon_0 A \frac{dE}{dt} = (8.854 \times 10^{-12})(0.030)(4.0 \times 10^{12}) = 1.1\ \text{A}$

Example 2: from a voltage ramp, with unit change

A 2.0 μF capacitor is charged so that its voltage rises at 1.5 × 10⁵ V/s. What is the displacement current between its plates?

Convert: C = 2.0 × 10⁻⁶ F.

$I_d = C\frac{dV}{dt} = (2.0 \times 10^{-6})(1.5 \times 10^{5}) = 0.30\ \text{A}$

The wires leading to the capacitor carry the same 0.30 A.

Example 3: circular plates

Round plates of radius 10 cm see the field change at 1.0 × 10¹³ V/(m s). Find $I_d$ .

Area: $A = \pi (0.10)^2 = 3.14 \times 10^{-2}\ \text{m}^2$

$I_d = (8.854 \times 10^{-12})(3.14 \times 10^{-2})(1.0 \times 10^{13}) = 2.8\ \text{A}$

Steps and units

Identify the surface the flux passes through, usually the area of the plates.
Find the rate of change: dE/dt directly, or dV/dt divided by the gap d, or dq/dt.
Multiply by $\varepsilon_0$ and the area, or use $C\,dV/dt$ .
Give the answer in amperes.

Check the unit: $\varepsilon_0$ in C²/(N m²) times flux rate in N m²/(C s) leaves C/s, which is the ampere. So displacement current has the same dimensional formula as ordinary current, [M⁰ L⁰ T⁰ I¹]. The calculator above uses the parallel plate form, with A in m² and dE/dt in V/(m s).

Conduction current and displacement current compared

Two kinds of current in a charging circuit
Feature	Conduction current	Displacement current
Cause	Moving charges, such as electrons	Changing electric flux
Where	Wires and conductors	Gaps, insulators, empty space
Exists for steady DC	Yes	No, it needs a changing field
Makes a magnetic field	Yes	Yes, in the same way

Adding $I_d$ to Ampere's circuital law makes it one of Maxwell's equations. A changing electric field then makes a magnetic field, just as a changing magnetic flux makes an emf (see the electric flux and magnetic flux pages). The two effects feed each other and travel as a light wave.

Common slips

Thinking charge jumps the gap. It does not; the dielectric or vacuum carries no conduction current.
Using the field instead of its rate. A large constant E gives zero $I_d$ .
Forgetting the area in cm². Convert 300 cm² to 0.030 m² before you multiply.
Fully charged capacitor. Once the voltage stops changing, both currents drop to zero.

In a dielectric filled capacitor, use

\varepsilon_r \varepsilon_0

in place of

\varepsilon_0

Frequently asked questions

What is the formula of displacement current?

It is I_d = ε₀ dΦ_E/dt, where Φ_E is the electric flux through the surface. For parallel plates this becomes ε₀A dE/dt, which equals C dV/dt for the capacitor.

Which unit is displacement current measured in?

The ampere, exactly like conduction current, because both describe a charge rate dq/dt. Only the I exponent survives, giving [M⁰ L⁰ T⁰ I¹].

Is displacement current a real current?

It is not a flow of charge, but it produces a magnetic field exactly like a real current does. In that sense it is just as real for magnetism.

Why was displacement current introduced?

Without it, Ampere's circuital law gives different answers for surfaces that pass through a wire and surfaces that pass through a capacitor gap. Adding I_d removes this contradiction and keeps current continuous.

Is displacement current zero in a DC circuit?

Once a DC circuit reaches steady state, the fields stop changing, so the displacement current is zero. It flows only while a capacitor is charging or discharging.