Displacement current — Maxwell's addition to Ampere's law: Worked examples

This is the practice arena for the parent topic. Before we solve anything, notice one habit that will save you every time: displacement current is never something new to memorise per problem — it is always one of a small number of case classes. Below we list every case the topic can throw at you, then we knock them all down one by one.

Two symbols you must already own before reading on (we earn them, per the contract):

$Φ_{E}$ = electric flux = "how much electric field pokes through a chosen flat area". Picture arrows of $E$ stabbing through a hoop; $Φ_{E}$ counts them. If $E$ is uniform, strength $E$ , and hits an area $A$ head-on, then $Φ_{E} = E A$ (V·m).
$ε_{0} = 8.85 \times 1 0^{- 12}$ (units C²/(N·m²)) = the permittivity of free space, a fixed number nature uses to convert "field growing" into "current". Think of it as an exchange rate.

The scenario matrix

Every displacement-current problem falls into exactly one of these cells. The worked examples are tagged with the cell they hit.

Cell	Case class	What you are given	Route to $I_{d}$
A	You know the flux rate directly	$d Φ_{E} / d t$	$I_{d} = ε_{0} d Φ_{E} / d t$
B	You know voltage ramp on a capacitor	$C, d V / d t$	$I_{d} = C d V / d t$
C	You know charge rate / wire current	$d Q / d t = I$	$I_{d} = I$ (they're equal)
D	Geometry: find $B$ between plates at radius $r$	$I, R, r$	$B = \frac{μ _{0} I r}{2 π R ^{2}}$
E	Edge / limiting: $r = R$ , or $r > R$ (outside)	boundary radius	switch formula at $r = R$
F	Zero / degenerate: constant field, DC steady state	$d V / d t = 0$	$I_{d} = 0$
G	Sign / direction: discharging capacitor	$d Q / d t < 0$	$I_{d} < 0$ → $B$ reverses
H	Real-world word problem	mixed clues	pick the matching row
I	Exam twist: field from area and $d E / d t$	$A, d E / d t$	$I_{d} = ε_{0} A d E / d t$

We now cover A through I with eight examples.

The worked examples

Worked example 4. (Cell D) The geometry classic —

$B$ between the plates Circular plates of radius $R = 5 cm$ carry a charging current $I = 2 A$ . Find the magnetic field $B$ at radius $r = 3 cm$ from the axis, in the gap. See the figure.

Forecast: is $B$ bigger or smaller than it would be around the full wire outside? (Hint: only part of the flux is enclosed.)

By symmetry $\oint B \cdot d l = B (2 π r)$ . Why this step? The field circles the axis like around a wire (look at the blue loop in the figure) — constant magnitude on a circle of radius $r$ .
Enclosed displacement current scales with area: $I_{d, e n c} = I \frac{π r ^{2}}{π R ^{2}} = I \frac{r ^{2}}{R ^{2}}$ . Why this step? The field between plates is uniform, so flux is proportional to area; the small loop of radius $r$ catches the fraction $r^{2} / R^{2}$ of the total.
Ampère–Maxwell: $B (2 π r) = μ_{0} I \frac{r ^{2}}{R ^{2}} \Rightarrow B = \frac{μ _{0} I r}{2 π R ^{2}}$ . Why this step? We equate circulation to $μ_{0}$ times enclosed (displacement) current and solve for $B$ .
Numbers: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 ) ( 0.03 )}{2 π ( 0.05 ) ^{2}} = 4.8 \times 1 0^{- 6} T$ .

Verify: the formula is identical to $B$ inside a uniform current-carrying wire, $B = μ_{0} I r / (2 π R^{2})$ — the promised symmetry ✓. Numerically $4.8 μ T$ .

Worked example 5. (Cell E) The edge and the outside — limiting behaviour

Same plates as Example 4 ( $R = 5 cm$ , $I = 2 A$ ). Find $B$ (a) exactly at the rim $r = R$ , and (b) outside the gap at $r = 8 cm$ . See the figure for the two regimes.

Forecast: does $B$ keep rising forever as $r$ grows, or does something switch over at $r = R$ ?

Inside ( $r \leq R$ ) we use Example 4's result. At $r = R$ : $B = \frac{μ _{0} I R}{2 π R ^{2}} = \frac{μ _{0} I}{2 π R}$ . Why this step? Set $r = R$ ; the two $R$ 's partly cancel. This is the peak — beyond the rim there's no more flux to enclose.
$B (R) = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 )}{2 π ( 0.05 )} = 8.0 \times 1 0^{- 6} T$ .
Outside ( $r > R$ ) the loop encloses the entire displacement current $I_{d} = I = 2 A$ (green shading in the figure), so $B (2 π r) = μ_{0} I$ and $B = \frac{μ _{0} I}{2 π r}$ . Why this step? Once $r > R$ you already caught all the flux; adding radius adds no more enclosed current, so $B$ now falls as $1/ r$ .
At $r = 8 cm$ : $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 )}{2 π ( 0.08 )} = 5.0 \times 1 0^{- 6} T$ .

Verify: at the rim both formulas must agree — the inside branch $μ_{0} I r / (2 π R^{2})$ at $r = R$ gives $μ_{0} I / (2 π R)$ , exactly the outside branch $μ_{0} I / (2 π r)$ at $r = R$ . Continuous, no jump ✓. And $B$ peaks at the rim, then decays (forecast confirmed) ✓.

Worked example 6. (Cell F) The degenerate case — fully charged, nothing changing

The same capacitor is now fully charged and sits at a constant voltage. The wire current is zero. Find $I_{d}$ and the magnetic field in the gap.

Forecast: with no current in the wire, do you expect any field in the gap?

$\frac{d V}{d t} = 0 \Rightarrow \frac{d Φ _{E}}{d t} = 0$ . Why this step? The field between the plates is fixed (charge no longer moving), so its flux is frozen.
$I_{d} = ε_{0} \frac{d Φ _{E}}{d t} = 0$ . Why this step? No change in flux means no displacement current — the effect exists only while the field is moving.
Hence $\oint B \cdot d l = μ_{0} (0 + 0) = 0$ , so $B = 0$ everywhere in the gap.

Verify: consistent with the wire — a compass near a fully charged, static capacitor stays still (forecast confirmed: no field) ✓. This is the "off" state that bookends every charging problem.

Worked example 7. (Cell G) Sign / direction — a

discharging capacitor A capacitor discharges through a resistor. Its stored charge falls as $Q (t) = Q_{0} e^{- t / τ}$ with $Q_{0} = 6 \times 1 0^{- 6} C$ and $τ = 0.5 s$ . Find the displacement current in the gap at $t = 0$ , and state its direction relative to the charging case.

Forecast: will $I_{d}$ come out positive or negative? What does the sign mean physically?

$I_{d} = \frac{d Q}{d t} = \frac{d}{d t} (Q_{0} e^{- t / τ}) = - \frac{Q _{0}}{τ} e^{- t / τ}$ . Why this step? Still $I_{d} = d Q / d t$ ; discharging just means $Q$ is decreasing, so the derivative is negative.
At $t = 0$ : $I_{d} = - \frac{6 \times 1 0 ^{- 6}}{0.5} = - 1.2 \times 1 0^{- 5} A$ . Why this step? Evaluate the exponential ( $e^{0} = 1$ ) at the start, where discharge is fastest.
The negative sign means the electric field between the plates is now shrinking instead of growing. So the induced $B$ circulates the opposite way compared with charging. Why this step? $B$ 's direction follows the sign of $d Φ_{E} / d t$ ; flip the sign, flip the curl.

Verify: magnitude $1.2 \times 1 0^{- 5} A$ ; sign negative. Units of $Q_{0} / τ$ : $C/s = A$ ✓. Reversed field matches the physics of a shrinking flux ✓.

Worked example 8. (Cells H + I) Word problem with a field-rate twist

The story: An engineer tests a capacitor in vacuum. She cannot measure current directly, but her sensor reports that the uniform electric field between two square plates of side $L = 10 cm$ is climbing at $\frac{d E}{d t} = 5 \times 1 0^{9} V/(m\cdots)$ . What displacement current is flowing across the gap?

Forecast: the field rate is huge — but $ε_{0}$ is tiny and the plates are small. Milliamps? Micro? Guess first.

Area of a plate: $A = L^{2} = (0.1)^{2} = 0.01 m^{2}$ . Why this step? Flux needs an area; the plates are squares, so $A = L^{2}$ , not $π R^{2}$ .
Flux $Φ_{E} = E A$ (field uniform and perpendicular), so $\frac{d Φ _{E}}{d t} = A \frac{d E}{d t}$ . Why this step? $A$ is a constant here — only $E$ changes — so it factors out of the derivative. This is Cell I.
$I_{d} = ε_{0} A \frac{d E}{d t} = (8.85 \times 1 0^{- 12}) (0.01) (5 \times 1 0^{9})$ . Why this step? Combine the definition with the flux we just built.
$I_{d} = 4.425 \times 1 0^{- 4} A \approx 0.44 mA$ .

Verify: units $\frac{C ^{2}}{N\cdotm ^{2}} \cdot m^{2} \cdot \frac{V}{m\cdots}$ . Using $V = N\cdotm/C$ gives $C/s = A$ ✓. Sub-milliamp — matches the forecast that tiny $ε_{0}$ tames even a giant field rate ✓. Note: no charges crossed the vacuum — pure Cell A/I physics, exactly what makes light possible.

Recall Self-check: match the clue to the cell

Given only $d V / d t$ and $C$ ::: Cell B → $I_{d} = C d V / d t$ Given only $d E / d t$ and plate area ::: Cell I → $I_{d} = ε_{0} A d E / d t$ Asked for $B$ at $r > R$ ::: Cell E → $B = μ_{0} I / (2 π r)$ Capacitor fully charged, DC ::: Cell F → $I_{d} = 0$ $Q (t)$ decreasing ::: Cell G → $I_{d} < 0$ , field reversed

Connections

Parent topic
Ampère's Law — the circulation rule we apply in every geometry example
Parallel Plate Capacitor — the canonical stage for Cells B–G
Gauss's Law — gives the uniform gap field used in the flux
Faraday's Law of Induction — the mirror twin for the sign discussion
Maxwell's Equations · Electromagnetic Waves · Speed of Light