Explain Displacement Current - QS Study
QS Study

We can use Gauss’ law to replace p = Ɛ0 Δ • E in the continuity equation to get

Δ. J + [ϐ (Ɛ0 Δ • E) / ϐt] = 0

So, Δ . (J + JD) = 0

Where, JD = Ɛ0 (ϐE/ϐt)

is the displacement current density, and has the same dimensions as the current density. Since JD is caused by a changing electric field, it will be non-zero between the capacitor plates.


Using Eq. [Δ . (J + JD) = 0], Ampere’s Law can then be modified to make Δ • (Δ x B) = 0, i.e.

Δ x B = μ0 J + μ0 Ɛ0 (ϐE/ϐt)

which is Ampere’s law as modified by Maxwell. From this we may conclude that currents and displacement currents are on an equal footing in electrodynamics, and that a changing electric field induces a magnetic field.

Returning to the problem of the charging capacitor, and assuming the electric field is uniform between the plates and zero elsewhere, Gauss’ law gives

E = (σ / Ɛ0) = (1/ Ɛ0) (Q/A)

where a is the surface charge density, Q is the charge and A the plate area. Thus between the plates the displacement current is


Hence, the total displacement current ID between the plates is identical to the current I in the wires charging the plates.