Say I have a circuit consisting of a battery, a wire, an open switch, and a capacitor. The circuit is open since the switch is open.

My book says that the capacitor will only be charged when the switch is closed, but I don't see why this is true. I would expect the capacitor to be charged a little - not as much as if the circuit is closed, but still charged none the less.

To further illustrate my point consider this: If the circuit is open, the current must be zero. Consequently the field must be zero. For the field to be zero, the capacitor's field must cancel out the battery's field. Therefore the capacitor must be charged.

Generalizing this concept, shouldn't capacitors be charged (to a lesser degree) in open circuits?

EDIT: In other words, if the field is zero, the capacitor must be charged to cancel out the field of the battery.

  • 1
    $\begingroup$ Generalizing for an open circuit, I think that the capacitor $may$ be charged. It could be discharged, fully charged, or anywhere in between. In some sense, it depends on what was happening right before the circuit became open. $\endgroup$ – Carser Mar 12 '14 at 2:51
  • $\begingroup$ @Jedediyah Why wouldn't it certainly be charged, since the switch can be considered another capacitor in series? $\endgroup$ – dfg Mar 12 '14 at 2:52
  • $\begingroup$ Ohh I see now, you mean that the circuit is open because there is a gap "inside" the capacitors. This technically doesn't count as an open circuit since that gap is part of the component in the circuit. So your example is of a closed circuit that definitely would charge the capacitors. $\endgroup$ – Carser Mar 12 '14 at 2:59
  • $\begingroup$ @dfg The premise of your question assumes that even in open circuit there is an electric field of the battery which is untrue. In a conductor without any current (electrostatic equilibrium) no electric fields can exist and hence there is no field which needs to be cancelled by the capacitor. The field due to batter is established only when the circuit is closed. If you want more elaboration, let me know. $\endgroup$ – Satwik Pasani Mar 12 '14 at 5:10
  • $\begingroup$ @SatwikPasani I understand that the net electric field must be zero in the wire. However, I don't see why the field produced by the battery is zero. The (charged) terminals would produce a field regardless of whether the circuit is closed or open. Imagine two parallel charged plates representing the terminals - they produce a field regardless of what's around them. So you would need something to cancel out the field right? $\endgroup$ – dfg Mar 12 '14 at 5:14

The capacitor will indeed be charged a little -- but the charge will be so low that we may as well call it uncharged. Here is why: the open switch is another capacitor (two conducting terminals, although not quite in plate form, separated by a dielectric). Its capacitance is extremely low, though: the terminals' cross section will be on the order of a $\mathrm{mm}^2$ rather than a few $\mathrm{cm}^2$ (for an electrolytic capacitor), the dielectric (air) has a much lower $\varepsilon$, and the distance between the terminals is of the order of a millimetre rather than in the micron range. The switch is in series with the (proper) capacitor, so their capacitances add reciprocally (i.e. $1/C = 1/C_\mathrm{Cap} + 1/C_\mathrm{Switch} \approx 1/C_\mathrm{Switch}$). Since total charge is $Q=UC$, it will be very low. This includes the part of the charge that goes to the switch, so the charge of the proper capacitor will be even lower.

  • $\begingroup$ This answer makes a lot of sense. I get that the switch has a very low capacitance, but is it really so low that the charge on the capacitor can be neglected? Sure, its nowhere as good of a capacitor as an actual capacitor, but the capacitance would have to be incredibly low for the charge in the capacitor to be neglected. $\endgroup$ – dfg Mar 12 '14 at 19:22
  • $\begingroup$ And also, what if the wire is disconnected from the terminal? Then the "open switch" can't really be considered a capacitor then can it? Because you don't have two conducting terminals separated by a dielectric. You just have 1 conducting terminal (the end of the wire) and the battery. $\endgroup$ – dfg Mar 12 '14 at 19:29
  • $\begingroup$ 1: a plate capacitor with $1 \mathrm{mm}^2$ plates at a distance of 1 mm (separated by air) has a capacitance of just below 0.01 pF. The smallest capacitors typically found in discrete circuits are larger by about a factor of 100, but even something in the nF range is not really a large capacitor. So, yes, the switch capacitance is very low. 2: If the wire is disconnected, that creates a third capacitor. It will probably be of still lower capacitance because the distance between wire and battery terminal will be much larger than in a switch, where the mechanics are well-controlled. $\endgroup$ – Ansgar Esztermann Mar 13 '14 at 10:47

You are right in principle, but by considering the field of a battery; you are considering something, which is considered negligible by your book. In real world problems, one always try to ignore effects which are negligible, to solve problems to a reasonable level of accuracy and highlight the principles.

If your book starts to talk of all these effects, it will also have to write, that the circuit is in a shielded cage, where no external fields can influence.

  • $\begingroup$ Sorry, I don't see why its negligible. The charge on the capacitor would have to be significant for the field to cancel out. $\endgroup$ – dfg Mar 12 '14 at 17:00

A capacitor in an open circuit $may$ be charged:

It could be totally discharged, or it could be that the switch was opened while the capacitor was fully charged. It is also known that capacitors leak. That is, they lose their charge over time.

This is a bit more of an argument about the semantics of your book, but I mostly agree with you. In your example (if I understand it correctly), the circuit isn't open because of the distances between the capacitors. That's a closed circuit, and in fact one that definitely (as you point out) charge the capacitors. If you're talking about a circuit such as:

----| |--- | |---
|               |
|               |
|               |

Then this is a closed circuit that will charge the capacitors. (sorry for the ascii circuit, the -| |- are capacitors, the MMM is a resistor, and the (-+) is a voltage source).

  • $\begingroup$ Sorry, I still don't understand why it won't always be charged. Why can't an open circuit be seen as 2 capacitors in series? $\endgroup$ – dfg Mar 12 '14 at 3:00
  • $\begingroup$ Thanks, but I'm not talking about two capacitors in series, I'm talking about an open switch and a capacitor which can be treated as two capacitors in series. $\endgroup$ – dfg Mar 12 '14 at 3:05
  • $\begingroup$ @dfg Does my beautiful ascii circuit clear things up? The circuit I think you are describing is actually closed. $\endgroup$ – Carser Mar 12 '14 at 3:05
  • $\begingroup$ So a circuit with an open switch is closed? Isn't that contradictory? $\endgroup$ – dfg Mar 12 '14 at 3:06
  • $\begingroup$ Hmmm, I must just not understand your example. Specifically, where did the switch come in? $\endgroup$ – Carser Mar 12 '14 at 3:07

Your argument is: If the circuit is open, the current must be zero. Consequently the field must be zero. For the field to be zero, the capacitor's field must cancel out the battery's field. Therefore the capacitor must be charged.

Replace the capacitor with a resistance. Following your argument in the same way, there must be a voltage drop at the resistor equalising the batteries field. For this a current has to flow. In an open circuit. It doesn't.

Therefore your argument stated above is wrong. In an open circuit the battery generates no electric field.

One could be nit-picking and say a circiut with a capacitor is never closed, but we have to model it somehow and live with the models limitations

  • $\begingroup$ Sorry, I don't follow. In the case of the resistor, the field generated by the voltage drop across the switch will cancel out the field of the battery. The switch would essentially act like a capacitor. $\endgroup$ – dfg Mar 12 '14 at 17:06
  • $\begingroup$ Exactly. Same happens using the capacitor $\endgroup$ – Lord_Gestalter Mar 13 '14 at 5:42

Here is a feature of capacitors which could conceivably have caused you to believe, against the advice of your (unreferenced) text book, that an uncharged capacitor with only one leg connected to a current source could be given a charge albeit as you admit a small one.

This answer takes a small number of steps and begins at wikipedia: https://en.wikipedia.org/wiki/Electrostatic_induction

At Wikipedia, scroll to the heading Induction in dielectric objects. It doesn't necessarily mean dielectrics in a capacitor, but in this case the nomenclature is exact.

To precis Wikipedia simply, a charge across the dielectric will distort the positioning of electron orbits around the nuclii of the atoms in the dielectric. When the charge of the conductors on each side of the dielectric is removed, the electron orbits in the dielectric atoms returns to normal. But not immediately.

My government's NZ Post Office Radio Training School (Wellington NZ, lecturer Grant,I. 1964), introduced this atomic distortion in capacitors which slowly return to normal when a DC charge was removed, by the term dielectric hysteresis (from my lecture notes). We students were told to short circuit capacitor terminals on any capacitor taken out of service which had been carrying high DC voltages, and leave the short circuit in place until the capacitor was re-installed.

Now to you. Did you by any chance build a practical circuit as described to experiment on before you asked your question, or are you being hypothetical as to the possibility of a "one-lead" capability to charge a capacitor because of some seemingly anomalous feature you may have observed in a DUT, particularly a high voltage circuit and under specific test conditions?

Here is my accidentally derived empirical data so you know where I am coming from: I removed a 4uF oiled paper capacitor which had been working at 5,000VDC from a valve transmitter. Left the terminals shortcircuited for 2-3 weeks. Removed the short for a subsequent 2 weeks. Thought I should replace the short before physically moving a very large and very heavy capacitor elsewhere, and with a flash and a bang the capacitor demonstrated that dielectric hysteresis rebuilt an estimated 5-10 percent of charge after weeks of short circuit.

A genuine, significant charge on a capacitor that had had "nothing" hooked up to its terminals for weeks.

Hopefully this answer may explain a possibility how you may have developed your hypothesis which negates the text book?


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