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In my physics textbook there is an example of using capacitor switches in computer keyboard:

Pressing the key pushes two capacitor plates closer together, increasing their capacitance. A larger capacitor can hold more charge, so a momentary current carries charge from the battery (or power supply) to the capacitor. This current is sensed, and the keystroke is then recorded.

That makes perfect sense, and is kind of neat. What I am curious about, is what happens to that extra charge afterwards. Is there some sort of discharge mechanism? I suppose, that would be also necessary to differentiate between single keystrokes and continuous depression (register stroke current, then register the discharge). What would happen to the capacitor if there was no such discharge mechanism, but its capacitance was suddenly reduced?

If capacitance is reduced, and the charge stays the same, then, according to $Q = C \Delta V_C$, the difference of potentials on plates of capacitor should increase and exceed that of a power supply thus reversing the current. Is that what is happening, and the keystrokes are recorded by sensing not only the existence of the current, but also its direction?

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2 Answers 2

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The capacitance is proportional to $\frac Ad$ the ratio of the plate area to the distance. When $d$ is decreased, the capacitance rises and the voltage would fall. If the capacitor is connected to a fixed voltage, it will draw current to restore the voltage. Then when $d$ is increased again, it feeds that current back into the power supply, again keeping the voltage constant.

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  • $\begingroup$ So suppose work done in pulling the plates apart is W then the stored energy in the cell is increased by W ? (I mean in case the capacitor was connected to a cell in a closed circuit intially which maintained a constant potential difference) $\endgroup$
    – user74370
    Commented Sep 3, 2016 at 13:32
  • $\begingroup$ Since the circuit is at a constant potential difference and the pulling apart of the capacitor plates reduces the capacitance,the energy stored in the capacitor also decreases. The energy lost by the capacitor is given to the battery (in effect, it goes to re-charging the battery). Likewise, the work done in pulling the plates apart is also given to the battery. So, if work done in pulling the plates apart by the external agent is W then the stored energy in the cell is increased both by W and by an amount equal to the lose in energy of the capacitor. $\endgroup$
    – Physicpsycho
    Commented May 31, 2017 at 5:43
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If a capacitor is connected in series with a battery, then the potential difference between the plates is fixed and equal to the voltage of the battery. Therefore, if the capacitance changes, then the charge on the capacitor plates must change as well in order to keep the potential difference between the plates constant.

During charging, the flow of current is such that charges are pulled off of one plate, say plate $A$, so that it obtains a net positive charge, and charges are deposited on the other plate, say plate $B$, so that it obtains a net negative charge. Charge flows around the circuit from $A$ to $B$.

During discharging, precisely the same thing happens but in reverse.

If the capacitor, however, is disconnected from the circuit, say after being charged to a particular potential difference, then the charge on the plates will remain fixed, and a change in capacitance (like moving the plates together) results in a change in potential difference precisely as you point out.

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  • $\begingroup$ Relating to the last paragraph: so change in capacitance in disconnected condenser would not result in any mildly catastrophic event, such as violent discharge, unless that change is so profound as to precipitate dielectric breakdown? $\endgroup$
    – theUg
    Commented Feb 21, 2013 at 23:56
  • $\begingroup$ Also, what is wrong with the word “electrode”? Seriously. $\endgroup$
    – theUg
    Commented Feb 21, 2013 at 23:56
  • $\begingroup$ @theUg Yes I would agree with your characterization of the disconnected capacitor. As for the terminology, feel free to change it back, but I just thought "plate" would be less confusing for those reading the question on the feed: en.wikipedia.org/wiki/Electrode $\endgroup$
    – joshphysics
    Commented Feb 22, 2013 at 1:07
  • $\begingroup$ @joshphysics So suppose work done in pulling the plates apart is W then the stored energy in the cell is increased by W ? (I mean in case the capacitor was connected to a cell in a closed circuit intially which maintained a constant potential difference) $\endgroup$
    – user74370
    Commented Sep 3, 2016 at 13:29

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