Been stuck on this for a while and haven't been able to make sense of it.

So let's say that we have a parallel plate capacitor with plate area, $A$, charging at a rate $i$. The gap between the plates is filled with a simple dielectric material and has a relative permittivity, $\epsilon_r$, with $\epsilon_0$ being the permittivity of free space. The electric field in the capacitor gap is given by:

$$ \textbf{E} = \frac{Q}{\epsilon_r\epsilon_0 A} $$

The electric displacement field is given by:

$$ \textbf{D} = \epsilon_r\epsilon_0\textbf{E} $$

and the subsequent displacement current density is given by:

$$ \textbf{J}_D = \epsilon_r\epsilon_0\frac{\partial \textbf{E}}{\partial t} $$

But this means that,

$$ \begin{align} \textbf{J}_D &= \frac {1}{A}\frac{\partial Q}{\partial t} \\ &= \frac{i}{A} \end{align} $$

But surely I've made a mistake here? This would imply that the displacement current density of a parallel plate capacitor would be independent of the material in the gap...which doesn't sound right.

Can anybody shed some light on my mistake here.

  • 1
    $\begingroup$ The displacement current was introduced so the the “current” in a series circuit would be the same everywhere including between the plates of a capacitor. $\endgroup$ – Farcher Sep 12 '18 at 12:14
  • $\begingroup$ @Farcher I believe, this, perhaps slightly expanded, should be the answer. $\endgroup$ – V.F. Sep 12 '18 at 12:23

You did not make a mistake!

In Ampere's circuital law a line integral is evaluated $\displaystyle\oint_L \vec B \cdot d\vec l$ where $L$ is a closed loop.
That closed loop defines the edge of an open surface $S$ as shown in the diagram below which is an annotated picture of a butterfly net.

enter image description here

The important thing to realise is the there is no condition placed on the open surface $S$ other than the fact that the closed loop defines its edge.

If the closed loop $L$ does not move but the surface $S$ does move the value of the integral $\displaystyle\oint_L \vec B \cdot d\vec l$ must stay the same.

Imagine a wire leading carrying a current $I$ leading to the plate of a capacitor $C$.

enter image description here

For surface $S1$ there is no problem in equating the line integral $\displaystyle\oint_L \vec B \cdot d\vec l$ to $\mu_0I$ as is required by Ampere's law.

However the problem is surface $S2$ where the is no current passing through it and yet the line integral has a finite value unchanged from that when surface $S1$ was considered.

Now introduce a displacement current $\displaystyle \frac {d}{dt}\int_{S2} \vec D \cdot d\vec S$ which is has exactly the same value as $I$ the current passing through surface $S1$.

The problem goes away and there is agreement with your analysis in that the displacement current is independent of the material in the gap.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.