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What exactly would happen in the ideal case where a capacitor were to discharge into a wire with zero resistance?

Like in here:enter image description here

There is still charge on one plate of the capacitor, which will try to go to the other plate to make $\Delta V = 0$ across the capacitor, so there has to be a current $I$.

But as more charge goes to the other side of the capacitor, the more repulsion these will cause to more coming towards them. So I am assuming current cannot just be constant?

I am starting from $$ I = C \frac{dV}{dt}, $$ but I know neither $V(t)$ nor $I$.

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  • $\begingroup$ Even a zero-restance wire will have some non-zero inductance. You can model it as having an inductor present in the circuit. $\endgroup$
    – BowlOfRed
    Commented Sep 29, 2017 at 0:43
  • $\begingroup$ It would help to have a switch, also of 'zero' resistance to consider 'before' and after the switch is thrown. To consider an initial amount of energy in the cap. $\endgroup$
    – docscience
    Commented Sep 29, 2017 at 1:09
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    $\begingroup$ I'm voting to close this question as off-topic because it is about a non-physical phenomenon. $\endgroup$
    – garyp
    Commented Sep 29, 2017 at 1:35
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    $\begingroup$ If we'd throw away questions about ideal situations ("non-physical phenomena"), there would not be much Physics left... $\endgroup$
    – stafusa
    Commented Sep 29, 2017 at 8:02

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Even if the Ohmic resistance is 0 so that no energy is lost to heat it is impossible to have a radiation resistance (impedance) of 0. So such a wire will be a 100% efficient antenna and all of the energy will go to electromagnetic radiation.

https://en.m.wikipedia.org/wiki/Radiation_resistance

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This is a capacitor $C$ plus inductor $L$ circuit.

The inductor is the single turn loop and it may be small but it can still be significant.

Possibly a good way to understand what happens is to see the effect of having such an inductance in the circuit and then seeing what happens as you make the inductance smaller and smaller.
You can at the same time add some resistance into the circuit and see what effect it has.

If there is no resistance in the circuit then the current/voltages in the circuit will undergo simple harmonic oscillation of frequency $f = \dfrac{1}{2 \pi \sqrt{LC}}$.

What is happening is that the energy originality stored in the capacitor is then stored in the inductor then back to the capacitor then back to the inductor etc.

What you need to do is solve a differential equation of the form

$L \dfrac{dL}{dt} + \dfrac QC = 0 \Rightarrow L \dfrac{d^2I}{dt^2} + \dfrac IC =0$

In fact even with no resistance the circuit will behave as though it is damped because the accelerating charge within the circuit will radiate electromagnetic waves and so there will a radiative resistance term.

You will note that as $L$ is and smaller and smaller the frequency of oscillations gets larger and larger.

Making $L=)$ you then get to a non-physical situation.

With resistance in the circuit the differential equation becomes

$L \dfrac{d^2I}{dt^2} + +R \dfrac {dI}{dt} +\dfrac IC =0$

and the relative values of the components in the circuit will determine the exact form of the oscillations provided that the circuit is under damped.

The existence of "parasitic" inductance in a circuit can cause a problem in some (high frequency) applications.

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So I am assuming current cannot just be constant?

In the context of ideal circuit theory (where $i_C = C \frac{dv_C}{dt}$ holds), the current through the capacitor is constant only if the rate of change of the voltage across is constant.

But an ideal wire has zero volts across for any (finite) current through and so the voltage across is constant.

But this a contradiction since you've stipulated that the capacitor is charged and so, there must be a non-zero voltage across which must decrease as the capacitor discharges.

In this context of ideal circuit theory, you've drawn a contradiction.

One might try the move of modeling the current through as an impulse (delta 'function') and the voltage across as a step (you'll need a switch added to the circuit to make this clear) but this cannot remotely resemble a physical result for obvious reasons.

Add a small resistance so that the capacitor discharges exponentially as usual and then look at what happens in the limit as the resistance goes to zero; the peak current is unbounded while the time constant goes to zero.

Eventually, before the current gets too large, you have to consider the inescapable inductance of loop of non-zero area as well as the radiation resistance.

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  • $\begingroup$ Inductance I understand, it'd be a LC circuit with a harmonic oscillator frequency, but what do you mean with radiation resistance? $\endgroup$
    – SuperCiocia
    Commented Sep 29, 2017 at 0:53
  • $\begingroup$ Speaking in ideal terms with inductance, no resistance you'll just capture the energy that was originally in the capacitor as an LC resonance that propagates forever. Your perpetual motion machine. If indeed the connection is ideally zero impedance then where does the energy go? - with zero inducatnce, a resonant frequency of infinity! Yes the trouble with idealisms they tend to blow up. $\endgroup$
    – docscience
    Commented Sep 29, 2017 at 1:08
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    $\begingroup$ @SuperCiocia, the oscillations will be damped because energy will be lost to radiation and so the loss can be modeled as a resistance. Take a look at, e.g., this $\endgroup$ Commented Sep 29, 2017 at 1:23
  • $\begingroup$ I used to do that all the time -- i.e. short a capacitor. In my case is was a large oil-filled HV capacitor used for smoothing in a 3000 volt power supply I built for my 1000 watt linear RF amplifier back when I was in high-school (early 1960s). This was a ham radio amplifier for my transmitter. I would use a big screwdriver across the terminals and sometimes the heat from the arc would melt the tip of the screwdriver. $\endgroup$
    – K7PEH
    Commented Nov 1, 2017 at 16:03
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Here we seem to have a classical misinterpretation. The circuit diagram - a shorted loaded capacitor - is already a contradiction. The lines in the diagram do imply there are wires or conductors involved, but that is not true. The black lines look very similar to wires, but actually it is only a mathematical "=" sign. For every load of the capacitor not being exactly zero, we have these equations: V_capacitor = u1 <> 0 and at the same time V_connection_line = u2 = 0

This is a contradiction, since the lines in the diagram (again, not to be confused with wires, supraconductors etc.) define u1 = u2, which is already a contradiction given by the circuit diagram. Every reasoning with real world wires having L, R or C, with radiation etc. has nothing to do with that contradiction of an "ideal case" right in the circuit diagram. It is equivalent to the definition of a single number that is both zero and not zero at the same time. In any real world the connection lines have to be replaced with a network consisting of L, C, R elements, so of course in real world that simple circuit does never exist - like shown, it must not exist. The same is true, if the connection lines are replaced with a switch and a second capacitor holding a different voltage when the switch is switched on, which is the famous "where does the energy go to? - capacitor phaenomenom". It's the same type of problem described in Can a capacitor be charged without having resistance in the circuit? , taking the equivalency between an ideal battery of 0 volts and an ideal shortcut into account.

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