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I'm curious about how to determine/calculate the charge on a parallel plate capacitor with unequal voltages applied to both sides. With a capacitor made of two plates with significantly different areas, from what I've read, you use the area of the plates that overlaps in the formula (along with the relative permittivity and the distance between the plates): (e*A)/d. This formula determines the mutual capacitance between the overlapping portion of the two plates, and the portion of the plates that doesn't overlap would have some self capacitance like charged sheets.

For a capacitor with voltage applied to one side, my understanding is that one plate will acquire some amount of charge because of self capacitance (it would act like an isolated conductor). What about the charge on a capacitor with unequal voltages at either plate/sheet (with plates/sheets of the same size)? Like if you had a 1000 volt capacitor connected to the positive of a 500 volt battery and the negative of a 100 volt battery to the other, so relative to ground, you had +250 volts at one end and -50 volts at the other (and you connected the other end of each battery to ground). Would the charge of each of the sides of the capacitor be the mutual capacitance times the voltage applied? But since voltage is a relative concept, would the voltage on each plate, and thus the charge be equal? Like would there effectively be 300 volts for both plates even though two unequal voltages were being applied (and so the charge could be unequal)? The concept confuses me and I'd appreciate some clarity.

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    $\begingroup$ Voltage is a difference. $\endgroup$ – user121330 Dec 5 '17 at 0:20
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You seem to believe that each terminal of a battery supplies half the voltage, and that the average of those two voltages is always 0V, or ground. This is wrong. Instead, you always have to think of what a voltage is measured relative to. Even “ground” is just some arbitrary choice that may not be the same in one place as it is in another.

In the case of a capacitor, all that matters is the voltage measured on one side, relative to the other side. That other side may be at ground, or it might be at a million volts; all that matters is the difference between the two sides.

Same thing with a battery: all that matters is the voltage measured at one terminal relative to the other terminal. For example, you might put two AA batteries in some little electronic device, with the top of one pressed to the bottom of the other. Each one on its own has a 1.5V difference between one end and the other. Pressing them together means there's no voltage difference between the positive of one and the negative of the other. So you get a total voltage difference 3V between one end of the pair and the other end. In principle, it doesn't matter that these are small voltages — the same idea applies to any voltage source. So if you have a 1000V source and a 1V source, you can — subject to certain conditions(*) — just connect the negative of one to the positive of the other and have a 1001V source.


(*) Note that this assumes that everything can handle whatever current ends up passing through this circuit, and that everything is nicely insulated. That last one is an important point, especially because insulators can change to conductors at high voltages. Basically, if you don't already understand these concepts very well, you shouldn't go trying them out with high-voltage sources. Stick to AAs if you want to experiment. They're cheaper and they won't kill you. I've got a scar from 5000V source I was working with a long time ago, and I understood all the principles well.

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  • $\begingroup$ With that being the case, does that mean that connecting the end of a battery made to supply 1 volt and one that can supply 1000 volts will be like connecting one 1001 volt battery to a capacitor? $\endgroup$ – Tom Dec 5 '17 at 1:30
  • $\begingroup$ Yes, that's true. See my updated answer for a little more explanation. And be careful if you're actually using high-voltage sources! $\endgroup$ – Mike Dec 5 '17 at 1:45
  • $\begingroup$ I appreciate the safety suggestion. While this does answer some of my question, what I’m still unclear about is can you give each part of a capacitor different charges with this scheme? Is the amount of net charge always limited by self capacitance? $\endgroup$ – Tom Dec 5 '17 at 2:09
  • $\begingroup$ @Tom, the charge is proportional to the potential difference between the two plates. The charges on the two plates are equal and opposite. $\endgroup$ – The Photon Dec 5 '17 at 3:07
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    $\begingroup$ Ah, I didn’t understand that part of your question. You have to bring a third conductor at a third voltage into the situation in order to even define “different” voltages on the two sides of the capacitor — you could call this ground if you want, or you could just call it $V_3$. Now, the two plates of the capacitor will have some capacitances relative to this third conductor. And if the plates are not symmetric with respect to the third conductor, they will have different charges. This kind of problem is not usually discussed at the elementary level. You can google “capacitance matrix”. $\endgroup$ – Mike Dec 5 '17 at 3:16

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