View Single Post
Nov17-08, 06:37 AM
Sci Advisor
HW Helper
tiny-tim's Avatar
P: 26,148
Quote Quote by Vanush View Post
Can someone please post a problem/situation outlining the motivation / usefulness of the electric displacement field? (Particularly with a capacitor) I'm having trouble understanding the concept and information is sparse on the internet.
Hi Vanush!

For electric displacement field in a capacitor, this (from the PF Library) may help:

Displacement current:

No current ever flows through a functioning capacitor.

But while a capacitor is charging or discharging (that is, neither at zero nor maximum charge), current is flowing round the circuit joining the plates externally, and so there would be a breach of Kirchhoff's first rule (current in = current out at any point) at each plate, if only ordinary current were used, since there is ordinary current in the circuit on one side of the plate, but not in the dielectric on the other side.

Accordingly, a displacement current is deemed to flow through the capacitor, restoring the validity of Kirchhoff's first rule:

[tex]I\ =\ C\frac{dV}{dt}[/tex]

and this linear displacement current [itex]I[/itex] (which might better be called the flux current or free flux current) is the rate of change of the flux (field strength times area) of the electric displacement field [itex]D[/itex]:

[tex]I\ =\ A\,\widehat{\bold{n}}\cdot\frac{\partial\bold{D}}{\partial t}\ =\ A\,\frac{\partial D}{\partial t}\ =\ C\frac{dV}{dt}[/tex]

which appears in the Ampére-Maxwell law (one of Maxwell's equations in the free version):

[tex]\nabla\,\times\,\bold{H}\ =\ \bold{J}_f\ +\ \frac{\partial\bold{D}}{\partial t}[/tex]

Note that the displacement alluded to in the displacement current across a capacitor is of free charge, and is non-local, since it alludes to charge being displaced from one plate to the other, which is a substantial distance compared with the local displacement of bound charge in, for example, the presence of a polarisation field.