A capacitor made of two parallel plates, each of area 'A' and separation 'd' is charged by an external dc sour

Question

Q. A capacitor made of two parallel plates, each of area 'A' and separation 'd' is charged by an external dc source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Ampere's circuital law, in its original form, links the magnetic field around a closed loop to the conduction current (moving charges) threading that loop:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c

Maxwell realised this law is incomplete. The classic illustration is a charging capacitor. Consider an Amperian loop encircling the wire that feeds one plate.

If you cap that loop with a flat surface cut by the wire, a real conduction current $I_c$ passes through it.
If you instead cap the SAME loop with a bulging surface that passes between the two capacitor plates, no charge crosses the gap — the space between the plates is an insulator. So $I_c = 0$ through this surface.

Ampere's law now gives two different answers for $\oint \vec{B}\cdot d\vec{l}$ for the same loop, depending on which surface you choose. That is a contradiction — the law cannot be right as it stands.

Maxwell's Fix: A Current Made of Changing Field

Between the plates there is no moving charge, but there is a growing electric field, because charge is piling up on the plates. Maxwell proposed that a changing electric flux acts like a current for the purpose of producing a magnetic field. He called it the displacement current, $I_d$ .

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

where $\Phi_E = \int \vec{E}\cdot d\vec{A}$ is the electric flux through the surface, and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space.

Check with the capacitor. For a parallel-plate capacitor of area $A$ and plate charge $q$ , the field between the plates is $E = \dfrac{q}{\varepsilon_0 A}$ , so the flux is $\Phi_E = EA = \dfrac{q}{\varepsilon_0}$ . Then

I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{1}{\varepsilon_0}\frac{dq}{dt} = \frac{dq}{dt} = I_c

So the displacement current in the gap is exactly equal to the conduction current in the wire. The two surfaces now give the same answer — the contradiction is gone.

The Complete (Ampere–Maxwell) Law

Maxwell rewrote Ampere's law so that the total current is conduction plus displacement current:

\oint \vec{B} \cdot d\vec{l} = \mu_0\left(I_c + I_d\right) = \mu_0 I_c + \mu_0\varepsilon_0 \frac{d\Phi_E}{dt}

Important

The deep meaning: a changing electric field produces a magnetic field, just as (by Faraday's law) a changing magnetic field produces an electric field. This symmetry is what makes self-sustaining electromagnetic waves possible — the changing E-field of the wave generates the B-field and vice versa.

Key Points to Remember …

Why this formula?

Displacement Current: Why the Formula Holds

The displacement current is one of the most elegant corrections in physics — it fixed a logical flaw in Maxwell's equations and predicted electromagnetic waves. Let's understand why its formula emerges.

1. The Problem That Demanded a Fix

Consider a capacitor being charged in a circuit. Ampère's law (in its original form) states:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

where $I_{\text{enc}}$ is the current passing through any surface bounded by the loop.

Now take two different surfaces bounded by the same loop:

Surface S₁: Cuts the wire — current $I$ passes through.
Surface S₂: Passes between the capacitor plates — no current passes through.

Surface	Current through it
S₁ (cuts wire)	$I$
S₂ (between plates)	$0$

This is a contradiction: the same loop gives two different values for $\oint \vec{B} \cdot d\vec{l}$ . Ampère's law is inconsistent for time-varying fields.

2. The Insight: Changing Electric Field

Between the capacitor plates, there is no conduction current, but there is a changing electric field as charge builds up.

The electric field between plates: $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$
As $Q$ changes, $E$ changes: $\frac{dE}{dt} = \frac{1}{\varepsilon_0 A} \frac{dQ}{dt}$

Maxwell realized: a changing electric field should produce a magnetic field, just like a current does.

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

The electric flux through the capacitor plates is:

\Phi_E = \int \vec{E} \cdot d\vec{A} = E \cdot A = \frac{Q}{\varepsilon_0}

Step 2: Differentiate with respect to time

\frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0} \frac{dQ}{dt} = \frac{I}{\varepsilon_0}

Step 3: Define displacement current

Maxwell defined the displacement current $I_d$ as:

I_d = \varepsilon_0 \frac{d\Phi_E}{dt}

From Step 2, this equals $I$ — the same conduction current in the wire. The displacement current "bridges" the gap.

4. The Corrected Ampère-Maxwell Law

The full law becomes:

\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_{\text{enc}} + I_d)

Or equivalently:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}

Why this works:

For surface S₁: $I_{\text{enc}} = I$ , $\frac{d\Phi_E}{dt} = 0$ → result = $\mu_0 I$
For surface S₂: $I_{\text{enc}} = 0$ , $\frac{d\Phi_E}{dt} = \frac{I}{\varepsilon_0}$ → result = $\mu_0 \varepsilon_0 \cdot \frac{I}{\varepsilon_0} = \mu_0 I$

Both surfaces give the same answer. The contradiction is resolved.

5. The Key Formula(e) — Summarized

Quantity	Formula	Meaning
Displacement current	$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$	Equivalent "current" from changing E-field

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The displacement current inside a charging capacitor equals the conduction current in the wires because the changing electric field between the plates produces a displacement current density that, when integrated over the plate area, exactly matches the rate of change of charge on the plates — which is the charging current itself.

Why This Works: The Core Idea

When a capacitor charges, charge flows onto the plates through the wires — that's the conduction current $I_c$ . But between the plates, there's no moving charge; the circuit appears broken. How can current be continuous? Maxwell resolved this by introducing the displacement current $I_d$ , which arises from a changing electric field in the gap. The key insight: the displacement current through the capacitor's interior is exactly equal to the conduction current in the wires, ensuring that the total current (conduction + displacement) is continuous everywhere.

Let's prove this step by step.

Step-by-Step Derivation

1. Set up the capacitor parameters

Consider a parallel-plate capacitor with plate area $A$ and plate separation $d$ . It is connected to a DC source that supplies a charging current $I_c(t)$ . At any instant, the charge on the positive plate is $Q(t)$ , and the potential difference across the plates is $V(t) = Q(t)/C$ , where $C = \epsilon_0 A / d$ (assuming vacuum or air between plates).

2. Relate the conduction current to the charge

The conduction current in the wire is simply the rate at which charge accumulates on the plate:

I_c(t) = \frac{dQ}{dt}

This is the current measured by an ammeter in the circuit.

3. Find the electric field between the plates

For a parallel-plate capacitor (ignoring edge effects), the electric field between the plates is uniform and given by:

E(t) = \frac{\sigma(t)}{\epsilon_0} = \frac{Q(t)}{\epsilon_0 A}

The field points from the positive plate to the negative plate. As the capacitor charges, $Q(t)$ increases, so $E(t)$ increases with time.

4. Define the displacement current density

Maxwell's displacement current density is defined as:

\mathbf{J}_d = \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}

For our parallel-plate geometry, the field is perpendicular to the plates, so the magnitude is:

J_d(t) = \epsilon_0 \frac{dE}{dt}

5. Compute the displacement current through the capacitor

The displacement current $I_d$ is the flux of $\mathbf{J}_d$ through any surface between the plates. Taking a surface parallel to the plates (area $A$ ), we get: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Displacement current is produced due to :$
(a) Constant electric field
(b) Constant magnetic field
(c) Changing electric field
(d) Changing magnetic field
›Reveal solutionSolution

(c) Changing electric field …

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CBSE 2026Set A1 markMCQ
$Q. The unit of displacement current is (A) Am (B) A (C) ΩA (D) ΩmA$
›Reveal solutionSolution

Displacement current is a form of current; its unit is the ampere (A).

Maxwell introduced the displacement current to complete Ampère's law:

$I_d=\varepsilon_0\frac{d\Phi_E}{dt}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. The concept of displacement current was given by a scientist named ______.$
›Reveal solutionSolution

Displacement current was proposed by James Clerk Maxwell to fix an inconsistency in Ampere's circuital law for circuits with a changing electric field (as between capacitor plates).

Maxwell noticed that the original Ampere's law failed for a charging capacitor (no conduction current flows between the plates, yet a magnetic field is still observed there). He resolved this by introducing an additional 'disp …

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CBSE 2025Set X11 markMCQ
$Q.'Ampere-Maxwell Law' is written as (symbols have usual meanings) :$
(a) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$
(b) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i + \varepsilon_0 \dfrac{d\phi_E}{dt}$
(c) $\oint \vec{B}\cdot \vec{dl} = \mu_0 i$
(d) $\oint \vec{E}\cdot \vec{dl} = -\dfrac{d\phi_B}{dt}$
›Reveal solutionSolution

(a) $\oint \vec{B}\cdot d\vec{l} = \mu_0 i + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$ . Maxwell added the displacement-current term $\mu_0\varepsilon_0\dfrac{d\phi_E}{dt}$ to Ampere's law. Option ( …

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CBSE 2024Set 55/1/11 markMCQ
$E_x = E_0 \sin(kz - \omega t)$
›Reveal solutionSolution
Displacement current exists wherever the electric field varies with time. Only Region I has an explicit time dependence ( $\sin(kz - \omega t)$ ), so displacement current exists only in Region I. The correct option is (A).

Concept and Intuition

Displacement current is not a current of moving charges — it is a term Maxwell added to Ampère's law to account for changing electric fields. The key idea is simple: wherever the electric field changes with time, there is a displacement current density given by $J_d = \varepsilon_0 \frac{\partial E}{\partial t}$ .

So the question reduces to: in which of these four regions does the electric field explicitly depend on time? A field that is constant in time, or one that depends only on position (like a static pattern), produces no displacement current.

Let’s examine each region carefully.

Step-by-step reasoning

Region I: $E_x = E_0 \sin(kz - \omega t)$ This is a travelling wave — the field depends on both position $z$ and time $t$ through the combination $kz - \omega t$ . Compute the partial derivative with respect to time:

$\frac{\partial E_x}{\partial t} = E_0 \cdot (-\omega) \cos(kz - \omega t) = -\omega E_0 \cos(kz - \omega t)$
This is non-zero (except at isolated instants). Therefore, displacement current exists in Region I.

Region II: $E_x = E_0$ This is a constant, uniform field — no dependence on $t$ at all.

$\frac{\partial E_x}{\partial t} = 0$
No displacement current.

Region III: $E_x = E_0 \sin kz$ Here the field depends only on position $z$ , not on time.

$\frac{\partial E_x}{\partial t} = 0$
No displacement current. (This is a static sinusoidal pattern, like a standing wave at a frozen instant.)

Region IV: $E_x = E_0 \cos kz$ …
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CBSE 2024Set A11 markMCQ
$\oint \vec{B}\cdot\vec{dl}$
(a) $\mu_0 I_c + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$
(b) $\mu_0 I_c$
(c) $\varepsilon_0 \dfrac{d\phi_E}{dt}$
(d) $\mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$
›Reveal solutionSolution

(a) $\mu_0 I_c + \mu_0 \varepsilon_0 \dfrac{d\phi_E}{dt}$ …

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CBSE 2024Set A1 mark
$Q. Fill in the blank with appropriate word: Variable electric field produce ______ current.$
›Reveal solutionSolution

Maxwell showed that a changing electric field acts as a source of magnetic field, just like a conduction current — this is called displacement current.

Maxwell noted that Ampere's circuital law (in its original form) was inconsistent when applied to a charging capacitor. To fix this, he proposed that a time-varying electric field between the capacitor plates also gives rise to a magnetic field, exactly as a conduction current does. He called this the displacement current, defined as:

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The formula for displacement current (I_d) is -$
(a) μ0 dφE/dt
(b) μ0 ε0 dφE/dt
(c) ε0 dφE/dt
(d) (1/ε0) dφE/dt
›Reveal solutionSolution

Maxwell introduced displacement current to make Ampere's law consistent for time-varying electric fields (e.g. between capacitor plates); it equals ε₀ times the rate of change of electric flux.

Maxwell showed that a changing electric flux $\phi_E$ through a surface is equivalent, in its magnetic effects, to a real conduction current. He defined the displacement current as:

$I_d = \varepsilon_0 \dfrac{d\phi_E}{dt}$

…

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CBSE 2023Set 55/1/11 markMCQ
$Q. In the process of charging of a capacitor, the current produced between the plates of the capacitor is :$
(a) $-\varepsilon_0\dfrac{d\Phi_E}{dt}$
(b) $-\dfrac{1}{\varepsilon_0}\dfrac{d\Phi_E}{dt}$
(c) $\varepsilon_0\dfrac{d\Phi_E}{dt}$
(d) $\dfrac{1}{\varepsilon_0}\dfrac{d\Phi_E}{dt}$ where symbols have their usual meanings.
›Reveal solutionSolution
The current between the plates of a charging capacitor is the displacement current, given by $\varepsilon_0 \frac{d\Phi_E}{dt}$ , which matches option (c).

Why This Question Matters

When a capacitor charges, no actual charge carriers flow across the gap between the plates. Yet a magnetic field is observed around the gap — as if a current were flowing. Maxwell resolved this paradox by introducing the displacement current, a term that accounts for the changing electric field between the plates. This is the key idea behind the question.

The symbols have their usual meanings: $\Phi_E$ is the electric flux through a surface between the plates, and $\varepsilon_0$ is the permittivity of free space.

Step-by-Step Reasoning

Recall Maxwell’s correction to Ampere’s law Ampere’s circuital law in its original form ( $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$ ) fails for a charging capacitor — because the enclosed current changes depending on the surface chosen. Maxwell added a term to fix this:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{conduction}} + I_{\text{displacement}} \right)$
where $I_{\text{displacement}}$ is the displacement current.

Define the displacement current Maxwell showed that a changing electric field produces a magnetic field just as a conduction current does. The displacement current through a surface is defined as:

$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$
Here $\Phi_E = \int \vec{E} \cdot d\vec{A}$ is the electric flux through the surface.

Apply to the region between capacitor plates Between the plates of a charging capacitor, there is no conduction current — no free charges move across the gap. However, the electric field between the plates is changing as charge builds up. Therefore, the current that appears between the plates is purely the displacement current:

$I_{\text{between plates}} = \varepsilon_0 \frac{d\Phi_E}{dt}$

Check the sign …
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CBSE 2022Set HE2171 mark
$Q. What is the reason for origin of Displacement current?$
›Reveal solutionSolution

Displacement current arises from a changing electric field, and was introduced by Maxwell to make Ampere's law consistent for circuits with capacitors.

Maxwell noticed that Ampere's circuital law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$ , gave inconsistent results for a charging capacitor: for an Amperian loop drawn as a surface passing between the capacitor plates (where no conduction current flows), the law would predict zero magnetic field, contradicting the case of a surface cutting the connecting wire. To resolve this, Maxwell proposed that a time-varying electric field between the plates itself acts as a source of magnetic field, just like a current does, and defined the displacement current $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$ (proportional …

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CBSE 2020Set 55/1/11 mark
$Q. Write the mathematical form of Ampere-Maxwell circuital law.$
›Reveal solutionSolution
Ampere-Maxwell law generalizes Ampere's law by adding Maxwell's displacement current term; it states that magnetic circulation equals the sum of conduction and displacement currents: $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)$ .

Why Maxwell had to fix Ampere's law

Ampere's original circuital law worked beautifully for steady currents: the line integral of the magnetic field around a closed loop equals $\mu_0$ times the current threading through. But it failed spectacularly when currents changed with time, particularly in situations like a charging capacitor where current flows in the wires but no charge physically crosses the gap between the plates.

Maxwell realized the flaw: a changing electric field between the capacitor plates should contribute to the magnetic field just as a real current does. He introduced the displacement current — not a flow of charge, but a changing electric flux that has the same magnetic effect. This insight unified electromagnetism and predicted electromagnetic waves.

The mathematical statement

The Ampere-Maxwell circuital law can be written in two equivalent forms:

Integral form:
$\oint_C \vec{B} \cdot d\vec{l} = \mu_0 \left( I_{\text{enc}} + \epsilon_0 \frac{d\Phi_E}{dt} \right)$
where:

$\oint_C \vec{B} \cdot d\vec{l}$ is the circulation of magnetic field around closed path $C$

$I_{\text{enc}}$ is the conduction current passing through any surface bounded by $C$

$\epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current, with $\Phi_E = \int \vec{E} \cdot d\vec{A}$ being the electric flux

Differential form:
$\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$
where:

$\nabla \times \vec{B}$ is the curl of the magnetic field …
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CBSE 2020Set 55/2/11 markMCQ
$Q. Displacement current exists only when (A) electric field is changing. (B) magnetic field is changing. (C) electric field is not changing. (D) magnetic field is not changing.$
›Reveal solutionSolution
Displacement current arises from a changing electric field, not a changing magnetic field. The correct option is (A).

Concept and Intuition

The idea of displacement current was introduced by James Clerk Maxwell to fix a logical gap in Ampère’s law. Ampère’s law originally said that a magnetic field is produced by a conduction current (moving charges). But consider a charging capacitor: between the plates, no charges flow, yet a magnetic field is still observed there. Something must be “acting” like a current in that gap.

Maxwell realised that what changes between the plates is the electric field — it builds up as charge accumulates. He proposed that a changing electric field itself generates a magnetic field, exactly as a current would. He called this effect displacement current. So displacement current exists only when the electric field is changing with time.

Watch out

A common mistake is to think displacement current relates to a changing magnetic field — that’s electromagnetic induction (Faraday’s law), not displacement current. The two are symmetric but distinct.

Step-by-Step Reasoning

Recall the definition of displacement current

Displacement current density is given by $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$ .

The total displacement current through a surface is $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ , where $\Phi_E$ is the electric flux.

Both expressions contain a time derivative of the electric field — so if $\vec{E}$ is constant, $\frac{\partial \vec{E}}{\partial t} = 0$ and displacement current is zero.

Examine each option

(A) electric field is changing → $\frac{\partial \vec{E}}{\partial t} \neq 0$ → displacement current exists. …
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Q.A capacitor made of two parallel plates, each of area 'A' and separation 'd' is charged by an external dc source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.

Why This Works: The Core Idea

Step-by-Step Derivation

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