# What's the dual of Faraday's law of induction?

Faraday's law of induction tells us that if we subject a coil to a changing magnetic field, a voltage will be found across the coil.

Duality suggests something similar should exist for capacitors: if they are subjected to a changing electric field, a current will be found.

Does this phenomenon exist? What's it called? Is there a simple experimental apparatus one might construct to observe it?

• Yeah, it's called a capacitor. Put a current into it and you will find a changing electric field between the plates. The converse also works. Jan 7 '14 at 13:08
• @user36129 I know what a capacitor is. I mention it in the question. "a capacitor" isn't a phenomenon like electromagnetic induction. Faraday's law of induction is to inductors as what is to capacitors? Jan 7 '14 at 13:18
• You mean Gauss's law? That seems like a trivial answer. Both Gauss and Faraday are incomplete though; they only talk about the E- and B-field, and leave the H- and D-field untouched. A full representation of all the laws is Maxwell's equations. Jan 7 '14 at 13:37
• Possibly related: electronics.stackexchange.com/questions/79302/… Jan 7 '14 at 13:56
• There is the Ampere-Maxwell law. The line integral of the magnetic field is dependent on the flux of the sum of the time derivative of the electric field and the current density. As user36129 said, if you take a fixed closed path and change the surface you use to compute the flux you will find either a current or a changing magnetic field (or both). Namely an example where the flux comes only from a changing E is when you put the surface inside a charging caacitor Jan 7 '14 at 14:02

The dual of Faraday's Law is Ampere's Law but, while Faraday's Law is fundamental to the physics of an inductor, Ampere's Law is not fundamental to the physics of a capacitor.

Now, it is true that, in circuit theory, the capacitor and inductor are duals:

$$i_C = C\frac{dv_C}{dt} \leftrightarrow v_L = L \frac{di_L}{dt}$$

However, we have to be more careful outside the context of circuit theory.

In physics, the fundamental relationship

$$Q = CV$$

clearly requires the existence of electric charge and an electric scalar potential due to a conservative electric field. This equation relates electric charge and electric scalar potential.

The closest we can get to a dual of this is

$$\Phi = LI$$

which relates magnetic flux and electric current. But magnetic flux is not the dual of electric charge.

The missing ingredient here is the hypothetical magnetic charge (magnetic monopole) which is the dual of electric charge.* Were magnetic charge $Q_m$ (measured in webers) to exist, it would be a source or sink of a conservative magnetic field (measured in amperes per meter) and there would be an associated scalar magnetic potential (measured in amperes).

We could thus relate magnetic charge and magnetic scalar potential with a magnetic "capacitance" measured in henrys.

Further, we could relate electric flux to magnetic current (measured in volts) with an electric "inductance" measured in farads.

To summarize, while electric flux and magnetic flux are duals, and changing magnetic flux is fundamental to the physics of an inductor, changing electric flux is not fundamental to the physics of a capacitor. Indeed, it is the electric field itself, not the electric flux, that is fundamental.

*Assuming magnetic charge exists, Maxwell's equations become

$$\nabla \cdot \vec D = \rho_e$$

$$\nabla \cdot \vec B = \rho_m$$

$$\nabla \times \vec E = - (\vec J_m + \frac{\partial \vec B}{\partial t})$$

$$\nabla \times \vec H = \vec J_e + \frac{\partial \vec D}{\partial t}$$

Yes its called displacement current and its the correction added by Maxwell to Ampere's Law.

I'm not 100% sure about this but I'm prepared to have a shot at it so be kind with your ability to down vote. If I've got it wrong I'll remove it.

Imagine two plates of a (bigger) capacitor with an alternating voltage across them. There will be an alternating electric field between the plates and if you placed a smaller capacitor within that field there will be a voltage developed across its terminals. It's self evident, given that the electric field has X volts per metre - the smaller capacitor will succumb to being "charged" to a voltage equivalent to the main electric field x the gap between the plates of the smaller capacitor.

But, because of "duality" I believe the smaller capacitor plates have to be shorted out - does that short now carry an alternating current that reflects the increase in electric field in the vicinity of its plates. I think it will and possibly that is the duality with an alternating magnetic field inducing voltage in an open circuit coil.

Fundamental to a capacitor is the electric field between its electrodes. When current/charges flow into or out of the capacitor the field strength changes. The opposite is also true: If the electric field inside the capacitor changes, charges will be forced out of or into the capacitor, resulting in current flow.

A simple experiment seems not too easy to be actually performed. However, the concept is trivial: Take a capacitor* and place it into a strong, external electric field. Voltage (and current) on the capacitor's connections will result until the electric field inside the capacitor fully compensates the externally applied field, which is when current stops until the external field changes again in strength or direction.

*) A capacitor with a known direction of its electrical field will of course be required, and it will have to have the external field applied in just this direction. I mention this because electrolytic capacitors commonly available as electronic parts are usually rolled up inside to save space and their electrical field is therefore basically zero to/from the outside. Besides, those physically small capacitors with relatively high capacity compared to simple parallel-plate capacitors used for experimentation have extremely high field strengths (Megavolts per Meter) and the external field would have to match that order of magnitude to achieve measurable results.