# Displacement Current Density of a parallel plate capacitor

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.

• 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. – Farcher Sep 12 '18 at 12:14
• @Farcher I believe, this, perhaps slightly expanded, should be the answer. – 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. 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 $C$.

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.