$\vec{E}$-and $\vec{B}$-fields in RCL circuit

Question

Consider a basic RCL-Circuit consisting of an ideal resistor, an ideal capacitor and an ideal inductor in series.

Assume that at time $t = 0$ the current is zero and the capacitor is charged with a charge $Q$.

• How do the $\vec{E}$- and $\vec{B}$-fields look like in dependence of the time? I want to know them as vector fields in the conductors and in open region around the entire circuit. An analytic expression would be great, but some plots would suffice.

• For the electric field I want to have the total electric field and the components due to changing magnetic field as well as due to surface charges.

• I am also interested in the limit $R \to 0$.

Especially the electric field is interesting since it seems to be neglected most of the time. For the answer of the analogous question for more simple circuits see:

• Galli, Goihbarg: Energy transfer in electrical circuits: A qualitative account
• Müller: A semiquantitative treatment of surface charges in DC circuits
• Jefimenko: Demonstration of the Electric Fields of Current-Carrying Conductors
• https://physics.stackexchange.com/a/62166/6581

^{Edit: In the first version of the question I also wanted to know how to derive everything from Maxwell equations. Since there are no responses up to now, and to focus more on one aspect, I split the question into two. This one of how the fields look like and if it is solved I will pose a second one about how to derive it from Maxwell equations.}

Edit 2: The usual high school presentation where only the magnetic field in the solenoid is considered (but no electric field) is clear on that level but incomplete. And especially the case of the electric field seems to be complicated.

I did some research and found papers about electric fields in solenoids but I don't get the information together to a complete picture. So a numerical simulation would be great which includes all relevant details. Here are the papers I found, perhaps this is helpful to write a good answer for someone who understands the papers better than I do:

• http://nsi.snu.ac.kr/php/download.php?board=treatise&idx=38
• http://www.hep.princeton.edu/~mcdonald/examples/solenoid.pdf

And of course the references in the cited papers.

At least I am looking for a qualitative account, which makes the direction of the fields clear. Consider for example Feynman lectures II picture 22-9. Feynman concludes with indirect arguments something about the line integrals $V_i$. However I want to get a direct picture of how the electric fields really are.

Answer 1

In the body of the question, you mentioned the limit of R approaching 0. Let's begin from there. In this case we have what is called an L-C Oscillating Circuit.

For convenience I will assume that the left plate of the capacitor has charge $q(t)$ such that $q(0) = Q$. Similar to the simple discharge of a capacitor, the upper plate of the capacitor begins to loose positive charge (I will speak about +ve charge and conventional current, instead of electrons and electron flow current).

Clearly, at any time $t$, current in the conductor is: $$i = - \frac{dq}{dt} $$The negative sign appears as the charge on the capacitor is decreasing with time, but current is positive. Induced emf $e$ in the inductor is given by $e=L\frac{di}{dt}$ (click here for explanation).

Applying Kirchhoff's law across the circuit, we find that $L \frac{di}{dt} = \frac{q}{c}$. This is a differential equation whose solution is $i=i_0 sin(\omega t + \phi)$ where $\omega = \frac{1}{\sqrt{LC}} $; $i_0$ and $\phi$ are constants. Note that this equation and the differential equation $L \frac{di}{dt} = \frac{q}{c}$, resemble the equation for simple harmonic oscillation, q taking the place of displacement, i instead of velocity, and $\frac{di}{dt}$ for acceleration.

The circuit does indeed oscillate, with the extreme position being the capacitor charged upto Q, but with opposite polarity.

Note that the direction of i, E, and B changes after a time, before returning to the original state. The period of this oscillation is $T=2\pi\sqrt{LC}$. The charge on the capacitor, current through the circuit both vary as sine functons. As for the various fields, The Electric field inside the conductor is 0, since we are assuming that the conductor is ideal. So, the electric fields around the conductor are radial and constant. There is a magnetic field due to current flowing through the inductor and the conductor. The field due to the conductor current is concentric and coaxial with the wire(assuming the wire is straight), with strength of $B=\frac{\mu_0 i}{4\pi r}(\sin\theta_1 + \sin\theta_2)$ where the angles are those the length of the wire subtend. This value varies sinusoidally, at a given point in space. As is shown in the gif, the direction of magnetic field in the interior of the inductor depends on the direction of current flowing through it. The Magnetic Field inside the inductor is given as $$B = \mu_0 n i=\mu_0i_0 \sin(\omega t) n$$ where n is number density of the coils. Similarly Electric Field between the plates of the capacitor will be $$ E=\frac{q}{A\epsilon_0} = \frac{Q \cos \omega t}{A\epsilon_0}$$ where A is the plate area of capcitor.

Adding a resistor to this setup is like providing friction to a simple harmonic oscillator. It serves only to dampen the oscillations. In short, After each oscillation, Charge on the capacitor will have been reduced. Also, a component of the Electric Field develops inside the resistor along the direction of current. It's magnitude is $E=\frac{J}{\sigma}$ where J is the current density in the resistor, and $\sigma$ the resistivity.

In such a case, a final steady state will be observed, where current and charge everywhere is 0. The differential equation becomes $$L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$$ You can find the solution for such an equation here

Answer 2

If your source term is the charged capacitor, I am imagining a switch in the system that you close at t=0, then you can find the current in the system.

Of course you care about the current in the system:

$$\oint \vec{B} \cdot \vec{dl} = \mu_0 (I_{Enclosed} + I_{Displaced}).$$ $$I_{Displaced}= \epsilon_{0} \frac{d}{dt}\Phi_E$$

And while the concerns above are valid I think the simplification that this is a parallel plate capacitor with a pseudo 2D circuit is a fine place to start.

Answer 3

How do the $\vec{E}$ and $\vec{B}$ look like in dependence of the time?

I think Hadrian Evan's animation is good, +1. But there's an issue: the field is the electromagnetic field. The electric field and the magnetic field are "are better thought of as two parts of a greater whole". You mentioned Jefimenko, see this quote: "...neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields. Therefore, we must conclude that an electromagnetic field is a dual entity always having an electric and a magnetic component simultaneously created by their common sources: time-variable electric charges and currents."

I want to know them as vector fields in the conductors and in open region around the entire circuit. An analytic expression would be great, but some plots would suffice.

They aren't two distinct fields. See this from Minkowski: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect". Also see this from Maxwell: "a motion of translation along an axis cannot produce a rotation about that axis unless it meets with some special mechanism, like that of a screw".

For the electric field I want to have the total electric field and the components due to changing magnetic field as well as due to surface charges.

The changing magnetic is the electric field, and vice-versa. To appreciate that imagine you move past an electron. You don't create a magnetic field for that electron because you moved. In similar vein get rid of the capacitor and the inductor, and you've only got a wire. Move some electrons down it and you observe a concentric magnetic field around the wire. But you didn't really create a whole new field. You merely created a mismatch between the electromagnetic fields of the metal ions and the electrons. Then if you twist your wire into a spiral you've reconfigured the concentric field and you've got your inductor. It works a bit like a flywheel actually, see this article.

Especially the electric field is interesting since it seems to be neglected most of the time.

I wish it was neglected more. See the Minkowski quote above. The field of the electron is the electromagnetic field. When you set down two electrons near to each other they move apart linearly. They are subject to linear electric force. This is the result of their two electromagnetic fields interacting. Neither electron has an electric field. The electric field between the plates of the capacitor is a place where you see this linear E-force only. The magnetic field around the wire or inductor is where you see the rotational B-force only, because there's some relative motion between you and some of the electromagnetic fields, like when you move past the single electron.

I did some research and found papers about electric fields in solenoids but I don't get the information together to a complete picture.

You only get the complete picture from the electromagnetic field. And like Maxwell and Minskowski said, it has this screw nature. Which is why we have electric motors. Shove the current along the wire, and the motor turns. It's a bit like a pump-action screwdriver. We also have generators where you turn the rotor to push current along the wire. That's a bit like an ordinary screwdriver. We don't have a right hand rule for nothing.

I am looking for a qualitative account, which makes the direction of the fields clear. Consider for example Feynman lectures II picture 22-9.

I'm a bit of a Feynman fan, but I think you need something more "fundamental" than that. Have a look at this: "In fact Richard Feynman complained ^{[citation needed]} that he had been taught electromagnetism from the perspective of electromagnetic fields, and he wished later in life he had been taught to think in terms of the electromagnetic potential instead, as this would be more fundamental."

However I want to get a direct picture of how the electric fields really are.

The electromagnetic field is not totally unlike the gravitomagnetic field.