Displacement Current

Q: What is displacement current?

Displacement current is the term id = ε₀ dΦE/dt, equal to ε₀ times the rate of change of electric flux through a surface. It is measured in amperes and produces a magnetic field exactly like an ordinary current, even though no charge flows through the surface, for example between the plates of a charging capacitor.

Q: Why did Maxwell introduce displacement current?

Ampere's law gave two answers for the magnetic field near a charging capacitor: one using a surface cut by the wire and zero using a surface passing between the plates. Maxwell added the displacement current, due to the changing electric field in the gap, so that every surface gives the same total current and the same field.

Q: Is displacement current a real flow of charges?

No. Displacement current is not a movement of charge; it is the rate of change of electric flux multiplied by ε₀. It is called a current because it has the unit ampere and is a source of magnetic field in the same way as conduction current. It can exist in a perfect vacuum.

Q: How is the displacement current in a capacitor related to the current in the wires?

They are equal at every instant. Between the plates id = CdV/dt = dQ/dt, which is exactly the conduction current in the connecting wires. So the total current, conduction plus displacement, is continuous around the circuit and Kirchhoff's junction rule holds at each plate.

Q: What is the magnetic field between the plates of a charging capacitor?

For circular plates of radius R with displacement current id, the field at distance r from the axis is μ₀idr/(2πR²) inside the plate region and μ₀id/(2πr) outside it. It is zero on the axis, largest at the edge and circles the axis.

Q: What are Maxwell's four equations?

They are Gauss's law for electricity, Gauss's law for magnetism (no magnetic monopoles), Faraday's law of induction and the Ampere-Maxwell law, which includes displacement current. Together with the Lorentz force they describe all of classical electromagnetism and predict electromagnetic waves travelling at 1/√(μ₀ε₀), the speed of light.

Q: What is asked about displacement current in NEET?

NEET usually asks one-line questions: the displacement current in a capacitor whose voltage changes at a given rate, the fact that it equals the conduction current in the wires, its formula and unit, the term Maxwell added to Ampere's law, and which Maxwell equation rules out magnetic monopoles.

Q: How is displacement current tested in JEE Main and Advanced?

JEE Main asks for the magnetic field at a point between circular plates (inside or outside the plate radius), displacement current in RC and AC circuits, and dimensional questions. JEE Advanced adds non-uniform or dielectric-filled gaps, leaky capacitors and combinations with charging-circuit calculus.

PhysicsElectromagnetic WavesFor JEE aspirants

Displacement current is the current-like term $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ produced by a changing electric field. Maxwell added displacement current to Ampere's law because the old law gave two different magnetic fields for a charging capacitor. No charge crosses the gap between the plates, yet $i_{d}$ there equals the current in the wires, so the total current is continuous and a magnetic field exists in the gap. This page covers the Ampere-Maxwell law, capacitor results, the field between the plates and Maxwell's equations, all frequent one-step questions in JEE Main and NEET.

On this page1Why Ampere's law failed2Defining

i_{d}

3Ampere-Maxwell law4Capacitor cases5

B

between the plates6Maxwell's equations7Revision map

Key Formulas - Quick Reference

★ Must learn Displacement current: $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ (unit ampere); with a dielectric of constant $K$ in the gap use $ε = K ε_{0}$
★ Must learn Ampere-Maxwell law: $\oint B \cdot d l$ $= μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ _{E}}{d t}$ $= μ_{0} (i_{c} + i_{d})$
★ Must learn Parallel-plate capacitor: $i_{d} = ε_{0} A \frac{d E}{d t} = C \frac{d V}{d t}$ , equal to the conduction current $i$ in the leads
Displacement current density $j_{d} = ε_{0} \frac{d E}{d t}$ ; a loop of radius $r < R$ between plates of radius $R$ encloses $i_{d} \frac{r ^{2}}{R ^{2}}$
★ Must learn Field between circular plates: $B = \frac{μ _{0} i _{d} r}{2 π R ^{2}}$ for $r \leq R$ and $B = \frac{μ _{0} i _{d}}{2 π r}$ for $r \geq R$
RC charging: $i_{d} = \frac{V}{R} e^{- t / R C}$ ; AC supply $V = V_{0} sin ω t$ : $i_{d} = ω C V_{0} cos ω t$ (leads $V$ by $\frac{π}{2}$ )
★ Must learn The same constants give the speed of EM waves: $c$ $= \frac{1}{μ _{0} ε _{0}}$ $= 3 \times 1 0^{8} m s^{- 1}$

1. Why Ampere's Law Needed a Fix

Ampere's circuital law, $\oint B \cdot d l = μ_{0} i$ , says the line integral of $B$ round a closed loop equals $μ_{0}$ times the current through any surface bounded by that loop. Apply it to a loop round the wire of a charging capacitor:

Figure 1: The same loop through

P

, two surfaces, two answers.

S_{1}

is crossed by the current

i

;

S_{2}

is crossed only by the growing electric field between the plates. Maxwell's fix: a changing electric flux counts as a current,

i_{d} = ε_{0} \frac{d Φ _{E}}{d t}

Flat surface $S_{1}$ : the wire pierces it, so $B (2 π r) = μ_{0} i$ and there is a field at $P$ .
Pot surface $S_{2}$ (same rim, bottom between the plates): no charge crosses it, so the old law gives $B = 0$ at the same point $P$ .
One point cannot have two fields, so the law is incomplete. The only thing crossing $S_{2}$ is the changing electric field between the plates.

2. Displacement Current

Find what the changing field between the plates is "worth" as a current (plate area $A$ , charge $Q$ ):

Field between the plates: $E = \frac{Q}{ε _{0} A}$ , uniform and confined to the area $A$ .
Electric flux through $S_{2}$ : $Φ_{E} = E A = \frac{Q}{ε _{0}}$ .
Differentiate: $\frac{d Φ _{E}}{d t}$ $= \frac{1}{ε _{0}} \frac{d Q}{d t}$ $= \frac{i}{ε _{0}}$ , so $ε_{0} \frac{d Φ _{E}}{d t} = i$ , exactly the current in the wire.

i_{d} = ε_{0} \frac{d Φ _{E}}{d t}

The displacement current through a surface is $ε_{0}$ times the rate of change of electric flux through it. It is measured in amperes and produces a magnetic field exactly like a conduction current, although no charge moves across the surface.

Figure 2: Current never stops at a plate. In the wires

i_{c} = i

i_{d} = 0

; in the gap

i_{c} = 0

i_{d} = i

. The total

i_{c} + i_{d}

is the same everywhere, so Kirchhoff's junction rule holds at each plate once

i_{d}

is included.

Conduction current

i_{c}

Flow of charges (electrons, ions) in a conductor. Exists for steady DC too. Causes Joule heating $i^{2} R$ . Obeys Ohm's law.

Displacement current

i_{d}

Rate of change of electric flux, $ε_{0} \frac{d Φ _{E}}{d t}$ . Zero if $E$ is steady. No charge flow, no heating. Exists even in vacuum.

Key idea

Outside the plates

i_{c} = i

i_{d} = 0

; between the plates

i_{c} = 0

i_{d} = i

. Their sum is continuous, so the magnetic field is the same whichever surface you choose.

3. The Ampere-Maxwell Law

Maxwell's generalisation: the source of $B$ is the total current $i_{c} + i_{d}$ through the surface bounded by the loop.

\oint B \cdot d l = μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ _{E}}{d t}

For steady currents and static fields $\frac{d Φ _{E}}{d t} = 0$ and the law reduces to Ampere's law.
In a real material both terms can exist in the same region (no medium is a perfect conductor or insulator).
A region with no conduction current can still have a magnetic field if $E$ changes there. This is what lets EM waves travel through empty space.
The magnetic field measured just inside the plates (point $M$ ) matches the field just outside (point $P$ ), as the law predicts.

Exam Trick

Never compute the flux if you know the current. For any capacitor in a circuit, the displacement current between the plates equals the conduction current in the connecting wires at every instant: $i_{d} = i = C \frac{d V}{d t}$ . Most NEET questions end in one line.

Quick Recall: tap to check

Is there a displacement current between the plates of a fully charged capacitor still connected to a battery?

No.

E

is constant, so

\frac{d Φ _{E}}{d t} = 0

and

i_{d} = 0

(and

i_{c} = 0

in the wires).

What are the SI unit and dimensions of

ε_{0} \frac{d Φ _{E}}{d t}

Ampere;

[A]

, the same as current.

Does Kirchhoff's junction rule hold at a capacitor plate?

Yes, if the displacement current is counted: conduction current into the plate equals displacement current out of it into the gap.

4. Displacement Current in Common Capacitor Cases

With $E = \frac{V}{d}$ and $C = \frac{ε _{0} A}{d}$ , the definition becomes $i_{d} = ε_{0} A \frac{d E}{d t} = C \frac{d V}{d t}$ . Use the form that matches the data:

Situation	Displacement current in the gap	Remember
field changing at rate $\frac{d E}{d t}$	$i_{d} = ε_{0} A \frac{d E}{d t}$	density $j_{d} = ε_{0} \frac{d E}{d t}$
voltage changing at rate $\frac{d V}{d t}$	$i_{d} = C \frac{d V}{d t}$	equals the lead current
charging through $R$ from a cell $V$	$i_{d} = \frac{V}{R} e^{- t / R C}$	largest at $t = 0$ , zero when full
AC supply $V = V_{0} sin ω t$	$i_{d} = ω C V_{0} cos ω t$	leads $V$ by $\frac{π}{2}$ ; $i_{r m s} = ω C V_{r m s}$
constant charging current $I$	$i_{d} = I$ , $\frac{d V}{d t} = \frac{I}{C}$	E grows linearly
dielectric ( $K$ ) fills the gap	$i_{d} = K ε_{0} \frac{d Φ _{E}}{d t}$	still equals $C \frac{d V}{d t}$ with the new $C$

Figure 3: Charging through

R

: the charge rises as

q = C V (1 - e^{- t / τ})

while the current in the wires, and so the displacement current in the gap, decays as

i_{d} = \frac{V}{R} e^{- t / τ}

. NCERT Example 8.1 (

C = 1 nF

R = 1 M Ω

V = 2 V

τ = 1 ms

i_{0} = 2 μ A

, and at

t = τ

i_{d} = 0.74 μ A

Figure 4: Capacitor on an AC supply,

V = V_{0} sin ω t

i_{d} = C \frac{d V}{d t} = ω C V_{0} cos ω t

, so the displacement current leads the voltage by

\frac{π}{2}

and equals the conduction current in the leads at every instant. NCERT Exercise 8.2 (

C = 100 pF

230 V

ω = 300 rad s^{- 1}

i_{r m s} = ω C V_{r m s} = 6.9 μ A

5. Magnetic Field Between the Plates

Take circular plates of radius $R$ with displacement current $i_{d}$ spread uniformly over the plate area. Apply the Ampere-Maxwell law to a circle of radius $r$ in the gap, centred on the axis ( $i_{c} = 0$ there):

By symmetry $B$ is tangential and the same all round, so $\oint B \cdot d l = B (2 π r)$ .
Inside ( $r \leq R$ ) the loop encloses the fraction $\frac{π r ^{2}}{π R ^{2}}$ of the flux, so the enclosed displacement current is $i_{d} \frac{r ^{2}}{R ^{2}}$ .
$B (2 π r) = μ_{0} i_{d} \frac{r ^{2}}{R ^{2}}$ gives $B = \frac{μ _{0} i _{d} r}{2 π R ^{2}}$ .
Outside ( $r \geq R$ ) all of $i_{d}$ is enclosed: $B = \frac{μ _{0} i _{d}}{2 π r}$ , the same as beside the wire.

Figure 5: Field between circular plates (

R = 5 cm

i_{d} = 0.2 A

). Inside,

B = \frac{μ _{0} i _{d} r}{2 π R ^{2}}

rises linearly; outside,

B = \frac{μ _{0} i _{d}}{2 π r}

falls like the field of a wire. Peak

B = 0.8 μ T

r = R

; at

r = \frac{R}{2}

and

r = 2 R

it is half of that,

0.4 μ T

Figure 6: Right-hand rule with

\frac{d E}{d t}

in place of current. Growing

E

into the page gives clockwise

B

; when the capacitor discharges

\frac{d E}{d t}

reverses and so does

B

. The field lines are circles about the axis both inside and outside the plate region.

Exam Trick

The capacitor gap behaves like a thick wire of radius $R$ carrying current $i_{d}$ : the field rises linearly inside, peaks at the edge, $B_{ma x} = \frac{μ _{0} i _{d}}{2 π R}$ , and falls as $\frac{1}{r}$ outside. Quick check: $B (\frac{R}{2}) = B (2 R) = \frac{B _{ma x}}{2}$ . On the axis, $B = 0$ .

Key idea

Between circular plates

B \propto r

up to the edge and

B \propto \frac{1}{r}

beyond it; the direction follows the right-hand rule with

\frac{d E}{d t}

playing the role of current.

JEE Advanced

Leaky capacitor. A charged capacitor whose dielectric (permittivity $ε$ ) conducts slightly (conductivity $σ$ ) discharges through itself. Inside the dielectric the conduction current density is $σ E$ and the displacement current density is $ε \frac{d E}{d t}$ . Charge conservation gives $ε \frac{d E}{d t} = - σ E$ , so $E = E_{0} e^{- t / τ}$ with $τ = \frac{ε}{σ}$ , and the total current density $σ E + ε \frac{d E}{d t}$ is zero at every point and instant. The Ampere-Maxwell law then gives $B = 0$ everywhere, although charge is flowing (Figure 7).

Figure 7: A charged capacitor with a slightly conducting dielectric discharges through itself:

E = E_{0} e^{- t / τ}

with

τ = \frac{ε}{σ}

. The conduction current density

σ E

and the displacement current density

ε \frac{d E}{d t}

are equal and opposite, so the total current, and hence

B

, is zero everywhere.

6. Maxwell's Equations

Maxwell's four equations, with the Lorentz force $F = q (E + v \times B)$ , contain all of classical electromagnetism.

Law	Equation	What it says
Gauss's law (electricity)	$\oint E \cdot d A = \frac{Q}{ε _{0}}$	charges are sources of $E$
Gauss's law (magnetism)	$\oint B \cdot d A = 0$	no magnetic monopoles; $B$ lines are closed
Faraday's law	$\oint E \cdot d l = - \frac{d Φ _{B}}{d t}$	changing $B$ produces $E$
Ampere-Maxwell law	$\oint B \cdot d l$ $= μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ _{E}}{d t}$	currents and changing $E$ produce $B$

Figure 8: The symmetry Maxwell completed. Faraday's law: changing

B

produces

E

; displacement current: changing

E

produces

B

. Each keeps the other going, which is exactly an electromagnetic wave.

The laws are now nearly symmetric (not fully: there are no magnetic charges). A changing $B$ makes $E$ and a changing $E$ makes $B$ , so the two can sustain each other and travel as an electromagnetic wave at $c = \frac{1}{μ _{0} ε _{0}}$ . Putting in the numbers, $\frac{1}{( 4 π \times 1 0 ^{- 7} ) ( 8.854 \times 1 0 ^{- 12} )}$ $= 2.998 \times 1 0^{8} m s^{- 1}$ , the measured speed of light, so light is an EM wave.

Exam Trick

Dimensions settle many options. $ε_{0} \frac{d Φ _{E}}{d t}$ is a current $[A]$ ; $μ_{0} ε_{0}$ has dimensions of $\frac{1}{speed ^{2}}$ , $[L^{- 2} T^{2}]$ ; $\frac{1}{μ _{0} ε _{0}}$ is a speed. Any option that fails this test is wrong.

Quick Recall: tap to check

Which Maxwell equation says magnetic monopoles do not exist?

Gauss's law for magnetism,

\oint B \cdot d A = 0

Which term did Maxwell add, and to which law?

μ_{0} ε_{0} \frac{d Φ _{E}}{d t}

, to Ampere's circuital law.

What was the most important prediction of Maxwell's equations?

Electromagnetic waves travelling at

c

= \frac{1}{μ _{0} ε _{0}}

= 3 \times 1 0^{8} m s^{- 1}

, so light is an EM wave.

Key idea

Displacement current completes the symmetry of electricity and magnetism and is the reason electromagnetic waves exist.

7. Solving Problems and Revision Map

Use the flowchart for any "field due to displacement current" question, then the mind map to revise the concept.

Figure 9: Three steps for every displacement-current field question: find

i_{d}

, decide how much of it the loop encloses, apply the Ampere-Maxwell law to a circle of radius

r

Figure 10: Mind map of this concept. Cover a branch, recall its three points, then check.

8. Solved Examples

Solved Example 1

A parallel-plate capacitor with circular plates of radius

1 m

has capacitance

1 nF

. At

t = 0

it is connected in series with a

1 M Ω

resistor across a

2 V

battery. Find the magnetic field at a point between the plates, halfway between the axis and the edge, at

t = 1 0^{- 3} s

Solution:

$τ = R C = 1 0^{6} \times 1 0^{- 9} = 1 0^{- 3} s$ , so $t = τ$ . Current in the leads (and $i_{d}$ in the gap): $i$ $= \frac{V}{R} e^{- 1}$ $= 2 \times 1 0^{- 6} \times 0.368$ $= 7.36 \times 1 0^{- 7} A$ .

A loop of radius $r = 0.5 m$ encloses $\frac{r ^{2}}{R ^{2}} = \frac{1}{4}$ of it: $i_{d, e n c} = 1.84 \times 1 0^{- 7} A$ .

$B$ $= \frac{μ _{0} i _{d, e n c}}{2 π r}$ $= \frac{2 \times 1 0 ^{- 7} \times 1.84 \times 1 0 ^{- 7}}{0.5}$ .

Answer: $B \approx 7.4 \times 1 0^{- 14} T$ , tiny but not zero.

Solved Example 2

A capacitor has two circular plates of radius

12 cm

5.0 cm

apart, and is charged by a constant current of

0.15 A

. Find (a) the capacitance and the rate of change of potential difference (b) the displacement current across the plates. (c) Is Kirchhoff's junction rule valid at each plate?

Solution:

(a) $C$ $= \frac{ε _{0} π r ^{2}}{d}$ $= \frac{8.854 \times 1 0 ^{- 12} \times π \times ( 0.12 ) ^{2}}{0.05}$ $= 8.01 \times 1 0^{- 12} F$ $= 80.1 pF$ .

$\frac{d V}{d t}$ $= \frac{i}{C}$ $= \frac{0.15}{8.01 \times 1 0 ^{- 12}}$ $= 1.87 \times 1 0^{10} V s^{- 1}$ .

(b) $i_{d} = C \frac{d V}{d t} = 0.15 A$ , equal to the charging current.

(c) Yes, provided "current" means conduction plus displacement current: $0.15 A$ of conduction current enters the plate and $0.15 A$ of displacement current leaves it into the gap.

Answer: (a) $80.1 pF$ , $1.87 \times 1 0^{10} V s^{- 1}$ ; (b) $0.15 A$ ; (c) yes, with $i_{d}$ included.

Solved Example 3

A capacitor of circular plates, radius

R = 6.0 cm

, has

C = 100 pF

and is connected to a

230 V

AC supply of angular frequency

300 rad s^{- 1}

. Find (a) the rms conduction current (b) whether conduction and displacement currents are equal (c) the amplitude of

B

3.0 cm

from the axis between the plates.

Solution:

(a) $i_{r m s}$ $= ω C V_{r m s}$ $= 300 \times 100 \times 1 0^{- 12} \times 230$ $= 6.9 \times 1 0^{- 6} A$ .

(b) Yes: $i_{d} = C \frac{d V}{d t}$ is exactly the conduction current in the leads at every instant.

(c) Peak current $i_{0} = 2 \times 6.9 μ A = 9.76 μ A$ . Inside the plates: $B_{0}$ $= \frac{μ _{0} i _{0} r}{2 π R ^{2}}$ $= \frac{2 \times 1 0 ^{- 7} \times 9.76 \times 1 0 ^{- 6} \times 0.03}{( 0.06 ) ^{2}}$ .

Answer: (a) $6.9 μ A$ ; (b) yes; (c) $B_{0} = 1.63 \times 1 0^{- 11} T$ .

Solved Example 4

A parallel-plate capacitor of capacitance

25 μ F

is being charged so that its potential difference rises at

4 V s^{- 1}

. The conduction current in the connecting wires and the displacement current between the plates are
(A) zero, zero
(B) zero,

100 μ A

(C)

100 μ A

100 μ A

(D)

100 μ A

, zero

Solution:

Answer: (C). $i$ $= C \frac{d V}{d t}$ $= 25 \times 1 0^{- 6} \times 4$ $= 1.0 \times 1 0^{- 4} A$ $= 100 μ A$ in the wires, and the displacement current in the gap is the same $100 μ A$ .

Solved Example 5

The electric field between the plates of a capacitor (plate area

0.01 m^{2}

) increases at

1.0 \times 1 0^{12} V m^{- 1} s^{- 1}

. Find the displacement current. What would it be if a dielectric of constant

K = 5

filled the gap and

\frac{d E}{d t}

were the same?

Solution:

$i_{d}$ $= ε_{0} A \frac{d E}{d t}$ $= 8.854 \times 1 0^{- 12} \times 0.01 \times 1 0^{12}$ $= 8.85 \times 1 0^{- 2} A$ .

With the dielectric, $ε_{0} \to K ε_{0}$ : $i_{d} = 5 \times 88.5 mA$ .

Answer: $88.5 mA$ in vacuum; $0.44 A$ with the dielectric.

Solved Example 6

Circular plates of radius

5 cm

carry a displacement current of

0.2 A

. Find the magnetic field between the plates at (a)

2.5 cm

(b)

5 cm

(c)

10 cm

from the axis.

Solution:

(a) Inside: $B$ $= \frac{μ _{0} i _{d} r}{2 π R ^{2}}$ $= \frac{2 \times 1 0 ^{- 7} \times 0.2 \times 0.025}{( 0.05 ) ^{2}}$ $= 4.0 \times 1 0^{- 7} T$ .

(b) Edge: $B$ $= \frac{μ _{0} i _{d}}{2 π R}$ $= \frac{2 \times 1 0 ^{- 7} \times 0.2}{0.05}$ $= 8.0 \times 1 0^{- 7} T$ (maximum).

(c) Outside: $B$ $= \frac{μ _{0} i _{d}}{2 π r}$ $= \frac{2 \times 1 0 ^{- 7} \times 0.2}{0.10}$ $= 4.0 \times 1 0^{- 7} T$ .

Answer: $0.4 μ T$ , $0.8 μ T$ , $0.4 μ T$ ; the fields at $\frac{R}{2}$ and $2 R$ are equal (Figure 5).

Solved Example 7

The quantity

ε_{0} \frac{d Φ _{E}}{d t}

, where

Φ_{E}

is electric flux, has the dimensions of
(A) charge
(B) current
(C) potential difference
(D) magnetic flux

Solution:

Answer: (B). $Φ_{E} = \frac{Q}{ε _{0}}$ for a closed surface, so $ε_{0} Φ_{E}$ is a charge and its time derivative is a current: $[A]$ . It is the displacement current.

Solved Example 8

5 μ F

capacitor with circular plates of radius

4 cm

is connected to a supply

V = 100 sin (1000 t)

volts. Write the displacement current and find the largest magnetic field anywhere between the plates.

Solution:

$i_{d}$ $= C \frac{d V}{d t}$ $= 5 \times 1 0^{- 6} \times 100 \times 1000 cos (1000 t)$ $= 0.5 cos (1000 t) A$ .

$B$ is greatest at the edge ( $r = R$ ) when $i_{d}$ is at its peak: $B_{ma x}$ $= \frac{μ _{0} i _{0}}{2 π R}$ $= \frac{2 \times 1 0 ^{- 7} \times 0.5}{0.04}$ $= 2.5 \times 1 0^{- 6} T$ .

Answer: $i_{d} = 0.5 cos (1000 t) A$ ; $B_{ma x} = 2.5 \times 1 0^{- 6} T$ at the rim.

Practice Questions

The displacement current in a $2 μ F$ capacitor is $1 mA$ . At what rate is its potential difference changing?Answer: $500 V s^{- 1}$
The field between plates of area $0.02 m^{2}$ changes at $5 \times 1 0^{11} V m^{- 1} s^{- 1}$ . Find $i_{d}$ .Answer: $88.5 mA$
Circular plates of radius $10 cm$ carry $i_{d} = 1 A$ . Find $B$ between the plates at $5$ , $10$ and $20 cm$ from the axis.Answer: $1.0$ , $2.0$ and $1.0 μ T$
A $1 μ F$ capacitor charges through $1 k Ω$ from a $10 V$ cell. Find $i_{d}$ at $t = 0$ and at $t = 0.69 ms$ .Answer: $10 mA$ and $5 mA$ ( $0.69 ms = τ ln 2$ )
Does a wire carrying a steady direct current have a displacement current inside it?Answer: No: the field in the wire is constant, so $\frac{d Φ _{E}}{d t} = 0$
At what rate must the field between $1 m^{2}$ plates change to give $i_{d} = 1 A$ ?Answer: $\frac{1}{ε _{0}}$ $= 1.13 \times 1 0^{11} V m^{- 1} s^{- 1}$
Which of Maxwell's equations would change if magnetic monopoles were discovered?Answer: Gauss's law for magnetism ( $\oint B \cdot d A$ would no longer be zero)

Common Mistakes to Avoid

Watch out

Thinking displacement current is a flow of charge across the gap. No charge crosses; it is a changing electric flux.
Using the full $i_{d}$ for a loop smaller than the plates. Only the fraction $\frac{r ^{2}}{R ^{2}}$ is enclosed.
Using $B = \frac{μ _{0} i}{2 π r}$ for a point between the plates with $r < R$ . That formula holds only for $r \geq R$ .
Writing $i_{d} = μ_{0} ε_{0} \frac{d Φ _{E}}{d t}$ . The $μ_{0}$ belongs to the law, not to the current.
Saying $B = 0$ between the plates because no current flows there. The displacement current produces the same field as the wire current.
Forgetting that $i_{d} = 0$ once the capacitor is fully charged in a DC circuit (the field stops changing).
Using $ε_{0}$ when a dielectric fills the gap. Use $ε = K ε_{0}$ , or simply $i_{d} = C \frac{d V}{d t}$ with the new $C$ .
Claiming Kirchhoff's junction rule fails at a capacitor plate. It holds once displacement current is included.

Frequently Asked Questions

What is displacement current?

Displacement current is the term $i_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ , equal to $ε_{0}$ times the rate of change of electric flux through a surface. It is measured in amperes and produces a magnetic field exactly like an ordinary current, even though no charge flows through the surface, for example between the plates of a charging capacitor.

Why did Maxwell introduce displacement current?

Ampere's law gave two answers for the magnetic field near a charging capacitor: one using a surface cut by the wire and zero using a surface passing between the plates. Maxwell added the displacement current, due to the changing electric field in the gap, so that every surface gives the same total current and the same field.

Is displacement current a real flow of charges?

No. Displacement current is not a movement of charge; it is the rate of change of electric flux multiplied by $ε_{0}$ . It is called a current because it has the unit ampere and is a source of magnetic field in the same way as conduction current. It can exist in a perfect vacuum.

How is the displacement current in a capacitor related to the current in the wires?

They are equal at every instant. Between the plates $i_{d} = C \frac{d V}{d t} = \frac{d Q}{d t}$ , which is exactly the conduction current in the connecting wires. So the total current, conduction plus displacement, is continuous around the circuit and Kirchhoff's junction rule holds at each plate.

What is the magnetic field between the plates of a charging capacitor?

For circular plates of radius $R$ with displacement current $i_{d}$ , the field at distance $r$ from the axis is $\frac{μ _{0} i _{d} r}{2 π R ^{2}}$ inside the plate region and $\frac{μ _{0} i _{d}}{2 π r}$ outside it. It is zero on the axis, largest at the edge and circles the axis.

What are Maxwell's four equations?

They are Gauss's law for electricity, Gauss's law for magnetism (no magnetic monopoles), Faraday's law of induction and the Ampere-Maxwell law, which includes displacement current. Together with the Lorentz force they describe all of classical electromagnetism and predict electromagnetic waves travelling at $\frac{1}{μ _{0} ε _{0}}$ , the speed of light.

What is asked about displacement current in NEET?

NEET usually asks one-line questions: the displacement current in a capacitor whose voltage changes at a given rate, the fact that it equals the conduction current in the wires, its formula and unit, the term Maxwell added to Ampere's law, and which Maxwell equation rules out magnetic monopoles.

How is displacement current tested in JEE Main and Advanced?

JEE Main asks for the magnetic field at a point between circular plates (inside or outside the plate radius), displacement current in RC and AC circuits, and dimensional questions. JEE Advanced adds non-uniform or dielectric-filled gaps, leaky capacitors and combinations with charging-circuit calculus.

Previous year questions on Displacement Current

1 question from past papers, each with a step-by-step solution.

JEE Main 2026 Jan 21 Shift 2, Physics Q21

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