Given this figure:
simulate this circuit – Schematic created using CircuitLab
I observe that Out is mirroring what's on the left side of the capacitor, effectively becoming a wire.
I know that if you apply the formula of the impedance of C1 makes sense. But how about from a physics standpoint?
Why is that?
It's important to understand the difference between linear circuit theory and physics. Many beginners attempt to use a SPICE-like circuit simulator to find truths about physics, without realizing what they're looking for is not there.
In linear circuit analysis, we assume that Kirchhoff's Voltage and Current Laws must always be true - voltage and current must exist in a closed loop. A circuit simulator generates the loop equations in matrix form and applies linear algebra to solve it. It makes no attempt to analyze the electric field, magnetic field, trapped charges on a piece of metal, etc.
Thus, what you attempted to simulate - an open-ended capacitor with nothing connected to it - is technically an illegal move. It's like division by zero, as far as linear circuit theory is concerned, this "circuit" is meaningless and cannot be analyzed. It's called a floating node - a point in a circuit where nothing is connected to it. By the rules, the voltage here is undefined.
simulate this circuit – Schematic created using CircuitLab
But why does the circuit simulator still shows an output? When the circuit simulator sees you're trying to simulate circuits with a floating node, it simply insert a huge virtual resistor to ground to force the formation of a closed loop, allowing the use of KVL and KCL again. Typically, the resistor is on the order of gigaohm to teraohm.
Why would you want to do this? In the real world, any voltmeter or oscilloscope has a large-but-finite input resistance. If you construct this open-ended capacitor circuit in the real world, the moment you put a voltage probe onto it, the loop is completed anyway. The circuit with a probe and the circuit without, are different physical systems. In most applications, only the first form has a practical purpose in engineering. The default behavior of inserting virtual resistance by a circuit simulator is essentially doing same thing.
In conclusion, what you're seeing here is just the voltage experienced by an extremely huge resistor, connected with a capacitor in series across a voltage source. There's nothing mysterious about it. You can even remove your 1 MΩ resistor R2, since the same result happens without any series resistor. The following circuit produces the same result in a simulator.
In one comment, the OP clarified that:
My question comes more from a physics standpoint
So the real question here is: what is physically happening on the metal plate at the open-ended side of the capacitor? What is its actual electric potential of that plate according to the laws of physics of Faraday and Maxwell? If its electric potential really is what the circuit simulator shows (even though the simulator does not simulate any electric field), why does it "act like a wire"?
In fact, this question has already been asked before. In What is physically happening when there is a square wave input on the left plate of a capacitor and open circuit on right plate of a capacitor?, its OP asked:
Why is it that the voltage transfers [across the capacitor], but it doesn't in [an open circuit] where Vout is 0 V. Is it related to the capacitors ability to store charge? If so, why does this matter as in the first picture the capacitor is not holding any charge as it's never charged due to a lack of current. And the lack of current is due to lack of a path for the current to flow to ground.
I'd like to inform you that these questions is one level deeper than circuit analysis, and most engineers are not equipped to answer your question.
Electronics operates at a model level that sits above where your question is at.
Your question is like asking why it is that 4 times 5 is 20. Most answers will form around the idea that adding up 4, 5 times, makes 20 or that adding up 5, 4 times, makes 20. None of the answers will dig deeper until you ask that question to a specialist of such questions, namely an abstract mathematician.
Electronics, at the level most try to gain it, assumes certain ideas into place. While they are true enough at the level needed, these devolve into circular arguments when you ask why they are what they are. Your question needs to be handled by a specialist of such questions, namely an experimental physicist. So this really isn't the right place unless you are lucky enough to find one here. I'm not one.
- jonk
He then attempted to explain it with a detailed answer from a physics-first perspective. See if it answers your problem. If not, I'd suggest you to try Physics Stack Exchange instead.
Your questions seems to have been inspired by our discussion in the comments to my post, https://electronics.stackexchange.com/a/649047/73158. As explained there, there is no conduction path between the two sides of the capacitor so the charge remains the same and since C is constant then from \$ V = \frac Q C \$ we can expect V to be constant too.
The impedance of a capacitor, in ohms, is given by 1/(2πfC).
In this case, C1 has an impedance of 1/(2 x π x 0.000047 x 1000). If you plug those numbers into a calculator, that works out around 3.4 ohms.
When compared with the 1M ohm resistor next to it, the capacitor has such a low impedance, that it's effectively a wire.
You can imagine a voltage divider consisting of the C, the R and the input impedance of your measurement setup. In the simulation the measurement setup has infinite impedance.
It's important to understand that resistors and capacitors (and inductors, transistors, etc.) are just models of physical/observable phenomena*. As such, we can redraw any circuit as long as it correctly models these phenomena (for example, a plate of metal by itself can act as a wire, but if you place another plate of metal in parallel, it can behave like a capacitor). Even air has a finite resistance (especially when it's humid). We can model air by adding an extremely large resistor** from \$V_{out}\$ to ground. Now we have a voltage divider with a transfer function near 1. From this perspective, if we were to build your circuit, the capacitor would act as a wire because it always forms a voltage divider in the physical world.
Of course, there may be other phenomena to model in the circuit --nearby components, wires, EMI, etc. All of these can have less-than-infinite (likely less than air) impedance paths to \$V_{out}\$ and hence influence it in ways that are undesirable --this is what is known as "floating" (while the AC voltage on \$V_{out}\$ will try to follow \$V_{1}\$, it will likely be noisy and easily affected by neighboring aggressors if not properly shielded).
As a side note, it seems you're wondering about this in the context of decoupling / filtering / interfaces. It's important to get familiar with how to model the input and output interfaces of circuits / amplifiers / ICs / etc. This may sound daunting, but it's as simple as remembering that every input and output interface passes signal imperfectly, and can be viewed as having an unwanted divider with whatever it touches (making signal transfer never exactly perfect, i.e. slightly less than 1). E.g. a voltage input, voltage output amplifier's input strives to have high input impedance (shorted to ground), whereas its output strives to have low output impedance (in series with the output) --but no matter what we do these will never be ideal (only acceptable). (of course, this would be different for a current input, current output amplifier). Again, every current or voltage based input/output has series/parallel impedance --it's up to us to remember which one will be relevant, i.e. what will affect the circuit negatively, so that we can design it well (ideally we want to pass signal with 0 attenuation between interfaces). Alluding back to your previous question, decoupling caps will never be in isolation, and will always form some sort of R[L]C filter. Another way to think of bypass caps is to consider them as discharge paths for unwanted signals with frequency above the signals of interest (for bypass caps, we only want DC, hence we put caps shunted to ground to form low pass filters).
*In the context of simulation/analysis. From the perspective of building a circuit, we use real materials with specific shapes in order to produce devices that are close to our models (i.e. that are useful).
**An un-changing ("time invariant") resistor may not be a good model of air; someone feel free to correct me. But the idea is that there's some less-than-infinite impedance between the floating node and ground.
Your previous question and this question are about the capacitor's basic property of maintaining a constant voltage across itself during rapid changes in current. Depending on how the capacitor is connected, we distinguish two main applications of this property:
1. Decoupling capacitor. In this application, one terminal of the capacitor is grounded and the input voltage source is connected in parallel, through a resistor, to the capacitor. Then, since its voltage is constant, the voltage at its other terminal is also constant. The RC integrating circuit from your previous question exploits this idea.
2. Coupling capacitor. In this application, the capacitor is connected in series (with one of its terminals) to the input voltage source. Since its voltage is constant, the (output) voltage at its other terminal follows the input voltage variations. AC amplifiers with capacitive coupling exploit this idea. There, coupling capacitors are charged to the bias voltage; so the input voltage is "shifted".
See also my answer to a similar question.
Because the circuit does not form a loop, as there is no load, just unconnected wire, it means that no current can flow in the circuit.
As no current flows in the circuit, there will be no current through the resistor and thus no voltage drop in the resistor. So voltage difference over resistor is always 0V.
As no current flows in the circuit, there is no current through the capacitor either. Thus there is no change of charge in the capacitor, so there is no voltage change over the capacitor. Assuming the capacitor was discharged to 0V to begin with, there will always be 0V difference over the capacitor.
Therefore, whatever voltage you set the capacitor left terminal with the AC source, no current flows anywhere and there will be 0V difference over capacitor and resistor so the output node has exact same voltage as the input node.
