2
$\begingroup$

If I had a circuit as shown in the figure with the battery and the capacitor being covered completely, save for a passage for the conducting wire, by a hypothetical material that doesn't allow electric and magnetic fields to propagate through it, an imaginary "energy blocking" material, would the capacitor get charged and store energy?

If we think in terms of the Poynting vector and flow of energy, the barrier would prevent energy transfer but if we think in terms of the charges being stored on each plate of the capacitor and the energy thereof then it seems as if energy will indeed get stored in the electric field of those charges.

Is this a paradox and if so then what is the apparent solution?

Or have i misunderstood the concept and implications of the Poynting vector and energy flux density?Circuit with Barrier

$\endgroup$

3 Answers 3

4
$\begingroup$

Easily solved: a true barrier slices the wires!

:)

If wires pass through, then it's not a barrier. (An energy-blocking barrier would be perfectly conductive. If wires touch it, a short circuit is created, and the E x B energy is reflected back to the battery.)

However, if the wires pass without contact through holes in the barrier, then the usual e-field and b-field are compressed into the small gap outside the wires (and the Poynting flux flows through the holes in the barriers.)

This is noted in JD Kraus Electromagnetics, where his barrier is a perfectly conducting sheet (which shorts out e-fields and reflects b-fields even down at zero frequency.) The E x B around the wire is "compressed" into the annular slot outside the wire yet inside the barrier-holes. The Poynting vector-field squeezes down small and dives into those annular slots, as in fig. b below.

This effect is found throughout E&M, where all circuits are waveguides, yet they can 'guide' waves with wavelength vastly larger than circuit dimensions. (100KM waves can fit inside 5mm coaxial cable! Amazing!)

Here's an odd concept which rubs our noses in the effect: an infinitesimal resonator, placed in a tiny pore in a perfect barrier, allows enormously wide EM waves to pass through, as if the tiny pore grows to ~lambda diameter. If placing resonators into tiny holes in a conductive plate is the high-frequency instance, then sending long columns of electrons through those holes is the DC case.

enter image description here

$\endgroup$
1
  • $\begingroup$ Did you try to quantify how much of the wave intensity gets through that much smaller-than-wavelength hole? Of course some will go through, but the expectation is that this will be very small part of the incoming wave intensity and most of it will reflect back. $\endgroup$ Aug 11, 2021 at 21:41
1
$\begingroup$

If the shield is blocking incoming EM energy, it has to block incoming electric charge as well, because two charged particles moving inside means that their EM energy moves inside.

If no charges are getting from outside to inside the barrier containing the capacitor, the capacitor plates can't get charged.

$\endgroup$
0
$\begingroup$

Perhaps there is an alternative? If the shield does not touch the wires and if the shield is kept a very large distance from the resistor.

The surface charge on the resistor and the current flowing through the resistor are unaffected. The $\mathbf E$ and $\mathbf B$ fields at the resistor are unaffected. The Poynting vector is unaffected. Nothing is affected. The concept of flow of energy through space from the battery through tiny holes to the resistor is simply not needed?

The flow of energy is a local effect at the resistor.

But I am no expert on the subject! Please ridicule my answer. Has anyone ever done an experiment to show there is no effect?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.