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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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

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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