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First, this question is different than Does an electrolytic capacitor have a minimum voltage? since that question is about circuit analysis.

So, my question is the following:

Let's pick two voltages, a high one and a low one. Let's say 5kV and 30V.

Let's imagine a capacitor made for a circuit operating at 5kV; if the dielectric is too thin, the charge may jump across; so the capacitor is designed to keep charges with a 5kV potential apart.

But if the dielectric is thick enough to handle higher voltages, how can a lower voltage, like 30V, attract a charge on the other side?

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    \$\begingroup\$ the electric field is always there. Like light. How can light shine so far? Is there a distance where it suddenly can't cross? \$\endgroup\$ – DKNguyen Aug 8 at 15:31
  • \$\begingroup\$ capacitors often use parallel foil plates, for efficient/dense energy storage. The parallel-plates provide a linearly-scalable medium in which to attract charge. Thus Q = C * V is the explanatory math. \$\endgroup\$ – analogsystemsrf Aug 8 at 15:48
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    \$\begingroup\$ There probably is a minimum, somewhere around the voltage required to attract a single electron to one plate (and not the other) within any required time constant. \$\endgroup\$ – hotpaw2 Aug 8 at 19:16
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But if the dielectric is thick enough to handle higher voltages, how can a lower voltage, like 30v, attract a charge on the other side?

Why not? The process is completely linear. There's no "threshold effect". I can use a capacitor rated at 50V to couple a signal that might be only a few µV in amplitude.

Also, it isn't the electric field that causes the electrons to move. It's the displacement of electrons (caused by an external source of voltage) that creates the field in the dielectric.

Suppose you have two capacitors of the same value, but one has 100× the dielectric thickness (and therefore 100× the area) of the other. If you charge them to the same voltage, they have the same charge — the same number of electrons have been shifted from one side to the other. Sure, the E field is 100× less intense in the one with the thicker dielectric, but there's also 10,000× as much (thickness × area) of it, so the total energy stored is also the same.

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  • \$\begingroup\$ Can you develop a little bit? I assumed the voltage would have a direct impact on the ability for the charge to cross the dielectric; and that a lower voltage would have a harder time attracting a charge through a wider medium. Is this wrong? \$\endgroup\$ – Thomas Aug 8 at 15:33
  • \$\begingroup\$ @Thomas It does. Thats why you get less charge at the same voltage with a lower capacitance. Q=CV. But that is less charge, not no charge \$\endgroup\$ – DKNguyen Aug 8 at 15:35
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    \$\begingroup\$ @Thomas I misread your post a few posts back. Same value cap charged to same voltage is same is and same charge and same time. rating doesnt matter as long as the cap survives. Two air tanks with two different max pressure ratings take the same amount of time to fill and hold the same amount of air at the same pressure. \$\endgroup\$ – DKNguyen Aug 8 at 15:54
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    \$\begingroup\$ "... and that a lower voltage would have a harder time attracting a charge through a wider medium. Is this wrong?" No, that is not wrong. For a given technology, a higher-voltage capacitor needs thicker dielectric; this makes the plates more lightly coupled, so they need to be larger. \$\endgroup\$ – TimWescott Aug 8 at 16:21
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    \$\begingroup\$ this makes the plates more lightly coupled, so they need to be larger just want to emphasize this point because it seems to be the part that @Thomas is missing. When you have your two theoretical capacitors, one high voltage and one low, with the same rating, the dielectric thickness is not the only difference - the higher voltage capacitor will also have larger plates - or else it wouldn't have the same capacitance value. \$\endgroup\$ – dwizum Aug 8 at 17:10

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