You are given a 2muF parallel plate capacitor. How would you establish an instantaneous displacement current o

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2\ \mu\text{F}

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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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Displacement current arises from a changing electric field. To get $1\ \text{mA}$ in a $2\ \mu\text{F}$ capacitor, you need to change the voltage at a rate of $500\ \text{V/s}$ — for instance, by connecting it to a source with $dV/dt = 500\ \text{V/s}$ .

Why Displacement Current?

In a capacitor, no actual charge flows through the insulating gap. Yet a changing electric field between the plates behaves exactly like a current — Maxwell called this the displacement current. It’s what makes the capacitor part of a complete circuit when the voltage changes.

The displacement current density is:

J_d = \varepsilon_0 \frac{dE}{dt}

For a parallel plate capacitor, the electric field is uniform: $E = V/d$ , where $d$ is the plate separation. So:

J_d = \varepsilon_0 \frac{1}{d} \frac{dV}{dt}

Multiply by plate area $A$ to get the total displacement current $I_d$ :

I_d = \varepsilon_0 A \cdot \frac{1}{d} \frac{dV}{dt}

But $\varepsilon_0 A / d$ is exactly the capacitance $C$ of a parallel plate capacitor. Hence the beautiful simplification:

I_d = C \frac{dV}{dt}

This is the key: displacement current equals capacitance times the rate of change of voltage. No need to track fields or geometry — just use $C$ .

Step-by-step

Identify the target. We need $I_d = 1\ \text{mA} = 1 \times 10^{-3}\ \text{A}$ . The capacitor has $C = 2\ \mu\text{F} = 2 \times 10^{-6}\ \text{F}$ .
Apply the relation. From $I_d = C \, dV/dt$ , solve for the required rate of voltage change:

\frac{dV}{dt} = \frac{I_d}{C} = \frac{1 \times 10^{-3}}{2 \times 10^{-6}} = 500\ \text{V/s}

Interpret the result. To sustain a displacement current of $1\ \text{mA}$ $1 mA$ , the voltage across the capacitor must be changing at exactly $500$ $500$ volts every second. This could be:
- A linearly increasing voltage: $V(t) = 500t$ (in volts, with $t$ in seconds).
- Or any waveform whose slope is $500\ \text{V/s}$ at the instant of interest — for example, a sinusoidal voltage $V(t) = V_0 \sin(\omega t)$ with $\omega V_0 = 500$ . …

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

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$60\ cm^2$
›Reveal solutionSolution
This tests Maxwell's displacement current for a charging parallel-plate capacitor: $I_d=\varepsilon_0 A\,\dfrac{dE}{dt}=\dfrac{\varepsilon_0 A}{d}\dfrac{dV}{dt}$ . Answer: $1.062\times10^{-2}$ A.

Concept and Intuition

Between the plates of a capacitor there is no conduction current, yet a changing electric field there produces a displacement current that seamlessly continues the circuit's current — this was Maxwell's key insight completing Ampere's law. For a parallel-plate capacitor, since $E=V/d$ , a changing voltage across the plates directly means a changing $E$ -field, and the displacement current works out to have exactly the same value as the conduction (charging) current feeding the plates, $I_d = C\,\dfrac{dV}{dt}$ .

Step-by-Step Solution

Displacement current: $I_d=\varepsilon_0\dfrac{d\Phi_E}{dt}=\varepsilon_0 A\dfrac{dE}{dt}=\dfrac{\varepsilon_0 A}{d}\dfrac{dV}{dt}$ .

Treat the source voltage of 400 V as being established (capacitor charged from 0 to 400 V) over the given interval $\Delta t=10^{-6}$ s, so $\dfrac{dV}{dt}\approx\dfrac{400}{10^{-6}}$ V/s.

Data: $A=60\ \text{cm}^2=60\times10^{-4}\ \text{m}^2=6\times10^{-3}\ \text{m}^2$ , $d=2\ \text{mm}=2\times10^{-3}\ \text{m}$ , $\varepsilon_0=8.85\times10^{-12}\ \text{F/m}$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Q. Which one of the following laws is modified by Maxwell to obtain four electromagnetic equations known as Maxwell's equations? (A) Gauss's law of electricity (B) Gauss's law of magnetism (C) Faraday's law (D) Ampere's circuit law$
›Reveal solutionSolution
Maxwell's key modification was adding the displacement current term to Ampere's circuital law, fixing its inconsistency for circuits with charging capacitors.

Concept and Intuition

Maxwell noticed that Ampere's original law, $\oint \vec B\cdot d\vec l=\mu_0 I_{enc}$ , gives inconsistent results depending on which surface bounded by the loop you use, when the current is discontinuous — for example, at a charging capacitor's gap, no conduction current crosses the surface between the plates. Maxwell resolved this by proposing a "displacement current" $I_d=\epsilon_0\dfrac{d\Phi_E}{dt}$ due to the changing electric flux, giving the corrected Ampere–Maxwell law:
$\oint \vec B\cdot d\vec l=\mu_0\left(I_{enc}+\epsilon_0\dfrac{d\Phi_E}{dt}\right)$
Gauss's laws (electricity and magnetism) and Faraday's law were left in their original forms; only Ampere's law needed this correction.

Step-by-Step Solution

Recall the four Maxwell's equations: Gauss's law (electricity), Gauss's law (magnetism), Faraday's law, Ampere–Maxwell law. …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$4.5\times10^7\ \mathrm{V\,m^{-1}\,s^{-1}}$
›Reveal solutionSolution
Displacement current in a capacitor equals $\varepsilon_0$ times the plate area times the rate of change of electric field, $I_d = \varepsilon_0 A\,\dfrac{dE}{dt}$ . With the given rate and plate radius, $I_d \approx 0.5\,\mu\mathrm{A}$ .

Concept and Intuition

Maxwell's displacement current arises from a time-varying electric field, exactly analogous to how a real conduction current arises from moving charge. Between the plates of a charging/discharging capacitor, $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} = \varepsilon_0 A \dfrac{dE}{dt}$ , where $A$ is the plate area. This displacement current is what "completes the circuit" through the gap between the plates in Ampère–Maxwell's law.

Step-by-Step Solution

Plate radius $r = 2\,\mathrm{cm} = 0.02\,\mathrm{m}$ , so area $A = \pi r^2 = \pi (0.02)^2 = 1.2566\times10^{-3}\,\mathrm{m^2}$ .

$I_d = \varepsilon_0 A \dfrac{dE}{dt} = (8.85\times10^{-12})(1.2566\times10^{-3})(4.5\times10^7)$ . …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$q = q_0\cos \omega t$
›Reveal solutionSolution
The displacement current through a capacitor equals the rate of change of the charge on its plates — just differentiate $q(t)$ .

Concept and Intuition

Maxwell's displacement current is defined precisely so that it equals the conduction current charging/discharging the capacitor: $i_d=\dfrac{dq}{dt}$ , ensuring current continuity even through the gap between the plates (where no charge physically flows).

Step-by-Step Solution

$q=q_0\cos\omega t$ .

$i_d=\dfrac{dq}{dt}=-q_0\omega\sin\omega t$ .

Common Mistakes …
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AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
$\epsilon_0 \dfrac{d\phi_\epsilon}{dt}$
›Reveal solutionSolution
This tests recognition of Maxwell's displacement current term in the Ampere-Maxwell law — it has the dimensions of current, since Maxwell explicitly introduced it as a current-like quantity to fix the inconsistency in Ampere's original law.

Concept and Intuition

Maxwell noticed that Ampere's law, $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$ , is inconsistent when applied to a capacitor being charged (no real current flows between the plates, yet a changing electric field exists there). He fixed this by adding a term $\epsilon_0\dfrac{d\phi_E}{dt}$ , calling it the displacement current $I_d$ , so that the corrected law reads $\oint \vec{B}\cdot d\vec{l} = \mu_0(I + I_d)$ . Since $I_d$ is added to an ordinary current $I$ inside the same equation, it must have the same dimensions as current — that is baked into how Maxwell constructed the term.

Step-by-Step Solution

Maxwell's displacement current is defined as $I_d = \epsilon_0\dfrac{d\phi_E}{dt}$ , where $\phi_E$ is the electric flux.

The generalized Ampere's law is $\oint \vec{B}\cdot d\vec{l} = \mu_0(I_{conduction} + I_d)$ . …
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AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$Q. The law which states that 'a variation in an electric field causes magnetic field', is (A) Faradays law (B) Bio-Savart Law (C) Modified Ampere's Law (D) Lenz's Law$
›Reveal solutionSolution
The statement describes Maxwell's correction to Ampere's Circuital Law (the displacement current), which predicts that a changing electric field generates a magnetic field.

Concept and Intuition

Original Ampere's Law related magnetic fields only to conduction currents. Maxwell noticed an inconsistency in circuits with capacitors (where no conduction current flows between the plates) and introduced the displacement current $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$ , showing that a changing electric flux itself acts as a source of magnetic field. This symmetric partner to Faraday's law (changing $B$ produces $E$ ) completes Maxwell's equations.

Step-by-Step Solution

Faraday's Law: changing $B$ field induces an EMF/electric field — not what's described here.

Biot–Savart Law: relates magnetic field to steady conduction current, not to a changing electric field.

Lenz's Law: only gives the direction of induced EMF, unrelated to this statement. …
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Q.You are given a $2\ \mu\text{F}$ parallel plate capacitor. How would you establish an instantaneous displacement current of $1\ \text{mA}$ in the space between its plates?

Concept understanding — Displacement Current

Displacement Current

The Problem Maxwell Spotted

Maxwell's Fix: A Current Made of Changing Field

The Complete (Ampere–Maxwell) Law

Key Points to Remember …

Why this formula?

Displacement Current: Why the Formula Holds

1. The Problem That Demanded a Fix

2. The Insight: Changing Electric Field

3. Deriving the Displacement Current Formula

Step 1: Relate charge to electric flux

Step 2: Differentiate with respect to time

Step 3: Define displacement current

4. The Corrected Ampère-Maxwell Law

5. The Key Formula(e) — Summarized

Why Displacement Current?

Step-by-step

Method: Relating Displacement Current to Capacitance and Rate of Voltage Change

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.You are given a 2 μF2\ \mu\text{F}2 μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA1\ \text{mA}1 mA in the space between its plates?

Why Displacement Current?

Step-by-step

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Q.You are given a $2\ \mu\text{F}$ parallel plate capacitor. How would you establish an instantaneous displacement current of $1\ \text{mA}$ in the space between its plates?

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