Is a capacitor in an open circuit charged?

Question

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.

Answer 1

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 the value of the effective capacitance would be lower than the lower value in the calculation, and would be very small). Since total charge across the capacitor-switch combination 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.

Answer 2

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.

Answer 3

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:

----| |--- | |---
|               |
|               |
|               |
----(-+)--MMM----

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).

Answer 4

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

Answer 5

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?