What form does the electric field take in a voltage difference?

Question

A laser of driving frequency $\omega$ emits a planar wave of the form $$\mathbf{E}(\mathbf{r},t) = Ae^{i(\mathbf{k} . \mathbf{r} - \omega t)}\hat{\mathbf{a}}$$ where $\mathbf{a}$ is the polarization of the electromagnetic wave.

In a voltage difference (a capacitor for instance), the amplitude of the electric field is given by $E = V/d$ where $d$ is the distance between the two plates, and the field is polarized in the direction of the two plates. Does this mean that the electric field takes the simple form $$ \mathbf{E} = \frac{V}{d} \hat{\mathbf{z}}$$ or is there any complex phase term, possibly with spatial and time dependance? I am particularly interested in Josephson junctions.

Answer 1

The potential difference in terms of the electric field is given as:

$$V=\int \vec{E}\cdot d\vec{s}$$

where $s$ is the distance from one end to another. In the case of the capacitor, since the electric field is constant inside the plates (and far away from the edges), you do get:

$$V=\int \vec{E}\cdot d\vec{s}=E\int ds=Ed$$

which is the result you have above.

On the other hand, your electric field has a clear dependence on the distance, so the potential difference will be different during the integration and indeed include complex phase terms. In the more general case, the polarization or the frequency can depend on the position, so you must take them into account for the integration.

For example, if the frequency doesn't vary with the distance (isotropic medium) and the polarization is parallel to the vector $d\vec{s}$, then the integral can be simplied to:

$$V=\int \vec{E}\cdot d\vec{s}=Aexp(-i\omega t)\int exp(i\vec{k}\cdot\vec{r})$$

and so on.