# Where is the electric field in an $LC$ circuit?

Suppose we have an Ideal $$LC$$ circuit with an open switch at time $$t=0$$ where the capacitor is initial charged to a voltage $$V$$. The instant the switch is closed, the typical $$LC$$ oscillations will begin. But what electric field actually causes these oscillations to begin? At time $$t=0+dt$$ the electric field in the circuit should appear as follows:

Note that there is no electric field inside the superconducting wires and similarly no field in the superconducting inductor. If this is the case, where is the electric field that supplies the initial force to get the electrons in motion? We know that $$F=ma$$ and electrons have a nonzero mass so they can only gain speed (they are initially at rest at $$t=0$$) if there is a non-zero electric field at their position. So where is the electric field that supplies this force?

My second issue is that there does not appear to be any location where an electric field can exist to supply the back emf. It is this back emf which must be overcome in order for the creation of a magnetic field within any inductor. That is, when we wish to create an magnetic field, we must do work against an electric field that opposes the motion of the charge and that energy supplied in opposition to the back emf materializes as energy within the magnetic field that we create. But there does not appear to be any field that could possibly provide that resistive back emf to oppose the motion of the charge of the current. Think about it this way: Imagine a positive charge moving from the positive plate to the negative plate as the capacitor discharges. In its traversal of the circuit, not once does it encounter an electric field to take away the energy it possesses and transform that energy into magnetic field energy? So then where is the location of the electric field providing the back emf?

Any help on this issue would be greatly appreciated as this issue has been driving me crazy!

The fact that the electric field in an ideal conductor is zero is not in conflict with the fact that the current in the wires is zero, but rather the opposite: even a negligible electric field in the conductor produces an arbitrary big current.

You can interpret your experiment in this way. Initially, there is an excess of electrons on the negative armor of the capacitor. Such electrons exert a repulsive force between each other, so when you close the circuit they will start to flow across the circuit towards the positive armor of the conductor. This flow of electron reduces the strength of the electric field, and hence, the energy, in the capacitor

If there were no inductance, the current generated in the circuit will be a Dirac's delta and in no time the charges will balance themselves out in the circuit.

However, the inductance will prevent this. The inductance will produce a voltage proportional to the time derivative of the current. If the current has a spike, you will have a very big potential across the inductor. This potential difference will limit the current flow across the circuit. Eventually, the potential difference along the inductance will be so high that it pushes the electrons back on the armor of the capacitor.

If you solve the differential equation of the circuit, you will see that the current has a harmonic functional behavior.

Alternatively, you can see this problem as the fact that the total energy oscillates between being stored in the electric field of the capacitor and being stored in the magnetic field of the inductance.

• Thanks for the response. Okay that all makes sense but I still have one issue: You state that "if the current has a spike, you will have a very big potential across the resistance" but the wires as well as the inductor are superconducting and have no resistance. Where is this potential/ electric field that acts as the back emf actually located if it can't be located in the wires or the inductor? – SalahTheGoat Apr 16 at 16:19
• @SalahTheGoat I wanted to write "if the current has a spike, you will have a very big potential across the INDUCTOR". I will edit my answer. – Davide Dal Bosco Apr 16 at 18:54

There was a field across the open switch. When the switch was closed, that field did not instantaneously disappear, but migrated. Zero field in an ideal conductor is true statically, but not dynamically. Transient fields can exist within an ideal conductor.

In ideal conductors (i.e., $$\sigma = \infty$$), a current can exist without any electric field pushing the charges along, as per Ohm's Law: $$\vec{J}=\sigma\vec{E}.$$ Furthermore, the back emf you talked about is really just the line integral of the E-field around the circuit loop. It has a nonzero value because an E-field permeates the space between the capacitor plates. Also, it's better to think about the energy in an LC circuit as being stored in the electric and magnetic fields, not in the charges. When the current is at a max (and the charge on the capacitor plates is zero), the energy stored in the B-field is also at a max. When the charge is at a max (and the current is zero), the energy stored in the E-field is at a max.

The capacitor acts as a spring. When the circuit is closed, the charges starts to spread through the circuit. The driving force is the repulsion between electrons, and that is the electric field.

The increase of the current, from its initial zero value, generates a voltage in the inductor, as the acceleration of a mass connected to the spring is proportional to the force of the string.

The outcome is an oscillating current and voltage.