Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the

IV. Numerical Problems · Q1

Q. Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.

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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

Displacement current is not a flow of charge; it is the effect of a time-varying electric flux.
It has the same units as ordinary current (ampere) and produces a magnetic field in exactly the same way.
It restores continuity of current: current is never truly broken, even across a capacitor gap.
Between capacitor plates $I_c = 0$ but $I_d \neq 0$ ; in a plain resistive wire $I_d \approx 0$ and $I_c$ dominates.

Note

With this correction the four Maxwell equations become fully consistent and predict that electromagnetic disturbances travel at speed $c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^8\ \text{m/s}$ — the speed of light.

Bottom line: the displacement current $I_d = \varepsilon_0\, d\Phi_E/dt$ is Maxwell's term that lets a changing electric field act as a source of magnetic field, completing Ampere's law and opening the door to electromagnetic waves.

Displacement current, introduced by Maxwell to fix Ampere's circuital law, is a defining concept of the NCERT Class 12 Physics chapter on electromagnetic waves, tested in CBSE boards, JEE Main and NEET. Anyone searching "displacement current definition and formula class 12 physics" will find this capacitor-gap explanation matches the NCERT-prescribed derivation of the Ampere-Maxwell law.

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
Ampère-Maxwell law	$\oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$	Magnetic field from both real and displacement currents
In differential form	$\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}$	Local version (for advanced study)

6. Why This Matters for Exams

Conceptual trap: Students often think displacement current is a real current of charges. It is not — it's a term that behaves like a current in producing magnetic fields.
Numerical problems: You'll often compute $I_d$ from $\frac{dE}{dt}$ or from the charging rate of a capacitor.
Key exam point: The displacement current is zero in steady-state DC circuits (constant fields), but non-zero in AC circuits or during charging/discharging.

7. The Deeper "Why"

The displacement current isn't just a mathematical patch — it reveals a profound symmetry:

A changing magnetic field produces an electric field (Faraday's law)
A changing electric field produces a magnetic field (Maxwell's correction)

This symmetry is what makes electromagnetic waves possible: each changing field sustains the other, allowing energy to propagate through empty space.

Final takeaway: The formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ holds because it makes Ampère's law consistent for all surfaces and reveals the deep symmetry between electricity and magnetism.

Step 1. Let the conduction current in the wire connected to the capacitor plates be $I_c=5$ A.

Step 2. As the plates charge, the charge on them changes at exactly the rate $I_c$ , i.e. $dq/dt=I_c$ ; the electric flux between the closely spaced plates is $\Phi_E=EA=q/\epsilon_0$ , so $\epsilon_0\,d\Phi_E/dt=dq/dt$ .

Step 3. The displacement current is defined as $I_d=\epsilon_0\,d\Phi_E/dt$ , so directly $I_d=dq/dt=I_c=5$ A -- this holds regardless of the plate radius R, since R cancels out of the relation between flux and enclosed charge.

Step 4. This confirms the consistency Maxwell's correction was designed to guarantee: whichever surface (through the wire or through the gap) is used to evaluate the current enclosed by the same Amperian loop, the total (conduction + displacement) current is identical.

✓Final answer

I_d=I_c=5

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