Magnetic field in a capacitor in a $LC$ circuit

Question

Consider a simple $LC$ circuit in which the charge on the capacitor varies sinusoidally. Current flows in this circuit, changing the charge on the capacitor and hence changing the electric field inside the capacitor. Shouldn’t this time varying electric field produce a magnetic field in the capacitor and similarly an electric field be produced in the inductor? I think this can create two problems :-

1. If we wish to calculate the total energy in the circuit we should technically be including the energy of the magnetic field in the capacitor and the electric field in the inductor right? However, we do not and instead conserve only the sum of the energies of the electric field inside the capacitor and magnetic field inside the inductor. This I don’t understand why.

2. Consider the capacitor :- A changing electric field induces a changing magnetic field which in turn induces a changing electric field and so on; it’s an infinite loop! How are we going to calculate the energies due to each and every field?

Answer 1

Textbook inductors and capacitors are idealizations. In real life, capacitors have inductance and inductors have capacitance. Design of capacitors and inductors is often influenced by the desire to minimize these "parasitics", so they may often be neglected.

The next level of complexity might associated a series inductance (typically a few nH) with a capacitor, and a parallel capacitance (typically a few pF) with an inductor.

Then, there are the interconnecting conductors. Those might be modeled as inductance and capacitance as well, or perhaps you need to use transmission line models.

For things like antenna theory, you may need to forget circuit theory altogether and build models of the electromagnetic field with appropriate boundary conditions where the conductors are.

This is just typical real world physics: you build models that capture the features of the problem that are important and neglect effects that are unimportant. You can generally imagine many more unimportant effects than important ones.

Answer 2

The ideal setting for this question is a double concentric sphere with radius $r$ and $r+dr$ as capacitor.

By linearity, the outer and inner fields for equal opposite charged spheres are zero, because the inner potential of a charged sphere is constant, the outer potentials are $\pm Q/(4 \pi r)$.

The inner and outer sphere are connected by a torus coil as inductor with outer wire length 0.

Capacitance and inductance so large and resistances everywhere so small, that the system oscillates for many periods.

Then the energy at any time is

$$\frac{1}{2} ( C\ U(t)^2 + L \ I(t^2)= \text{const}$$

The energy can be expressed by the volume integrals over $E^2$ in the capacitor and $B^2$ in the inductor or by double integrals over all pair products of charged areas with the Coulomb law of distance and the dipol interaction of all wire elements in the coil.

But one has to use only one of these picture consistently: Pair interactions of charges and dipols, field energy in space between, or total voltage and current multiplied by the constants C and L emerging in the integrals and depending on geometry only.

To show their equalities is the art of transformations of integrals over vector fields, named after Green, Gauss and Stokes.

Answer 3

The question, if the the B^2 field between the spheres should be included: Of course, its equal to $\dot E_r$ and its value proportional to the volume, linear in $dr$. Since its value is zero in a maximum $\dot U =0$ and the induced $B$-field has a maximum at $U =0$, we can be sure, that the formalism includes all fields in all volumes, even in the dynamic and open geometry cases.

But the proof - excluding radiation by the limits $ \omega \to 0, \lambda/r \to \infty$ - needs heavy artillery, at least in the conventional picture of time dependent vector fields in euclidean geoemetry in a rest system.

The geniusses of topology and cohomology, starting with Poincarés ideas, found that its quite easy in the quasistatic case:

There is the exterior algebra of differential forms generated by $d$ with the rule

Coordinate differentials $$\mathbb d: x \to dx$$ $$ \mathbb d dx =0,\quad \mathbb d a \wedge \mathbb d b =0, \text{if}\quad a,b \quad \text{linear dependent}$$ $$ A \to d A =F$$ Maxwwell I
$$ddA=dF=0: \quad \text {div}\ E=0, \quad \text{curl}\ B+ \dot E=0$$ Hodge dual by Lorentz scalar product with the 4-volume form $$*F = \left< dx^n,F\right>.$$
Maxwell II $$d*F=*j \quad \text{div}\ E = \rho\quad \text{curl}\ B -\dot E = j$$ Solution using Fourier transform generating the Coulomb kernel $$ d^{-1} *j = \int \frac{j(\xi)}{x-\xi} d^n\xi $$

This is pure alegbraic abstraction, but is becoming extremely helpful in case that:

1. The charges are concentrated in conductors in thermal equilibrium, such that, at any time, the volume integrals of $E$ can be replaced by a surface integral over the conductors with constant boundary values $U_i$. $E=0$ in the inner of the conductor. This is the low frequency, low resistance limit. This is the easy part of the game, because it deals with the scalar potential.

2. The current density at the surface of the conductors equals the tangent component of $B$ everywhere, while the tangent $E$ is zero. The volume integrals over $B$ can be replaced by surface integrals over the conducting wires with constant current over any cross area betwenn connction points.

This is the point were intuition of vector field integral breaks down, but in special relativistic 4-d exterior caluculus there is no other problem then to trust the diffential-integral algebra over intuition.

From life long experience we know, that about 3/4 of the students disappear from the lessons in this very moment.

Answer 4

When charge builds up across a capacitor, and the E flux through it increases, there is indeed an induced magnetic field around the capacitor, like there would be through a current carrying wire. If rate of E flux change (the current) changes, for example if the power source's voltage drops, the capacitor can act a tiny bit like an inductor would in steadying & smoothing the current drop. But this inductance is very tiny, until like GHz frequencies its affect on the capacitor is negligent. So yes, in theory everything capacitor can be more accurately modeled in circuits (lumped element modeling) by a capacitor in series with a very small inductor, maybe in the picoHenry scale. For practical purposes, this can be ignored in engineering until GHz frequencies.

Part of the energy indeed radiates away as you mentioned, with E fields producing B fields and so on. As a result, it takes slightly more energy to get a capacitor to a certain voltage than otherwise. This effect is small and for practical purposes can be ignored. It's like having an additional tiny resistance in series with the capacitor.

Those repeated inducements of fields in space don't continously need more energy from the origin. They just slush the energy forward like a disturbance in water propagates the original energy onward.

But the original energy is indeed not free, but a price to be paid by the circuit's current - it's as though there was an additional tiny resistance in the circuit, with the effect of reducing I for the same V. But again this effect is tiny as a capacitor doesn't radiate much energy outward. It's like a tiny pebble dropping into a pond rather than a big boulder.

Answer 5

Here is my answer to Q2. Faraday's law states that the time derivative of the electric field is the same thing as the rotation of the magnetic field. Ampere's says that the time derivative of the magnetic field is the same thing as the rotation of the electric field. All up to a sign and some constants. There is no infinite loop.