Displacement Current

PHYOrigin & Nature

Maxwell introduced the displacement current $I_d=\varepsilon_0\,d\Phi_E/dt$ so that Ampere's law would stay consistent in the gap of a charging capacitor, where no conduction charge crosses. Numericals ask you to find $I_d$ from a changing field $dE/dt$ , a changing voltage $dV/dt$ , or capacitor geometry.

The key idea is that between the plates a varying electric field acts exactly like a real current of equal magnitude.

Displacement current (flux form)

I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt} = \varepsilon_0 A \dfrac{dE}{dt}

A

= plate area (m

^2

E

= field between plates (V m

^{-1}

\varepsilon_0=8.85\times10^{-12}\ \text{F m}^{-1}

. Uniform field assumed.

From rate of change of voltage

I_d = C \dfrac{dV}{dt}

C

= capacitance (F),

V

= potential difference across plates (V); this equals the charging conduction current

I_c

in the wire.

Circular plates

I_d = \varepsilon_0 \,(\pi R^2)\,\dfrac{dE}{dt}

R

= plate radius (m), so area

A=\pi R^2

; convert cm to m before substituting.

In a charging capacitor the conduction current $I_c$ in the wire and the displacement current $I_d$ in the gap are equal, $I_c=I_d$ , which keeps the total current continuous.
$I_d=\varepsilon_0 A\,(dE/dt)$ and $I_d=C\,(dV/dt)$ are the same quantity expressed differently; use whichever data the problem gives.
Displacement current depends on the rate of change of the field, not its instantaneous value — a constant (even large) field gives zero $I_d$ .
It has the same dimensions as electric current and the SI unit ampere, identical to conduction current.
It produces a magnetic field exactly as a conduction current would, completing the symmetry $\oint\vec B\cdot d\vec l=\mu_0(I_c+I_d)$ .
Inside an ideal parallel-plate gap the field is taken uniform, so $\Phi_E=EA$ simplifies the flux derivative.
For circular plates remember $A=\pi R^2$ and convert the radius from cm to m (a common factor-of- $10^{4}$ error in area).

Where the marks go

Forgetting the $\varepsilon_0$ factor — $I_d$ is tiny because $\varepsilon_0\sim10^{-12}$ ; leaving it out inflates the answer enormously.
Not converting plate area or radius to SI (cm $^2$ to m $^2$ , cm to m) before substituting.
Confusing $dV/dt$ with $dE/dt$ : use $I_d=C\,dV/dt$ for voltage rate, $I_d=\varepsilon_0 A\,dE/dt$ for field rate.
Thinking a steady field or steady charge gives a displacement current — only a time-varying field does.

How the board asks it

Numerical $I_d=C\,dV/dt$
A parallel-plate capacitor of capacitance $100\ \mu F$ is being charged so that the potential difference across it increases at a steady rate of $10^5\ V\,s^{-1}$ . Calculate the displacement current in the gap between the plates.
Numericalcircular plates, $A=\pi R^2$ , $I_d=\varepsilon_0 A\,dE/dt$
The circular plates of a capacitor each have radius $5\ cm$ . If the electric field between the plates changes at the rate $2\times10^{12}\ V\,m^{-1}\,s^{-1}$ , calculate the displacement current between the plates.
Define / stateflux form $I_d=\varepsilon_0\,d\Phi_E/dt$
Define displacement current and write its mathematical expression in terms of the rate of change of electric flux.
Give reasons $I_c=I_d$ keeps total current continuous
While a parallel-plate capacitor is charging, no charge crosses the gap between its plates, yet a magnetic field exists there. Give reasons.
Derive / prove $\oint\vec{B}\cdot d\vec{l}=\mu_0(I_c+I_d)$
Explain why Maxwell modified Ampere's circuital law and obtain its modified form including the displacement current term.
Assertion–Reason $I_d$ depends on $dE/dt$ , not on instantaneous field
Assertion: A capacitor carrying a large but constant charge has zero displacement current between its plates. Reason: Displacement current depends on the rate of change of electric field and not on its instantaneous value. Choose the correct option regarding the two statements.

Practise this topic

NextField Relation E0 = cB0

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.