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I'm trying to understand the relationship between the capacitance with the voltage output of this circuit. If the capacitance is greater, why does it take more time to charge the plates of the capacitor? (Creating the "charge oposition" that manifests itself on the voltage "cut" seen in the simulation.)

If the capacitance is greater, I assume either the area of the capacitor plates is larger or the distance between the plates is smaller. Intuitively, how does the "larger areas" and the "smaller distance between the plates" affect the electrons' movement and the charge so that it manifests on charging velocity?

What differs between the simulations is just the capacitance:

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    \$\begingroup\$ For the same reason a larger bucket takes longer to fill. \$\endgroup\$ Sep 17 at 18:03

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You are asking about two different things.

First why does a bigger capacitor take longer to fill?

It's in the name of the thing -- the bigger a capacitor is, the more electrons it has to hold to come up to a certain voltage. Since you're charging it through a fixed resistor, the current vs. voltage relation of the charging circuit doesn't change -- but keep in mind that current is the speed of charge exchange, and the voltage vs. charge relationship of the capacitor does change.

Hence, longer charging for bigger caps, just like the "bigger bucket" analogy in the comments.

Second what makes a capacitor "bigger" (in the sense of more capacity).

  • If you take an electron away from a positive charge, it develops a voltage. The more the charges are separated, the higher the voltage is. So the voltage per charge of a capacitor goes up as the plates get more separate*, and the capacitance goes down.
  • If you put a bunch of electrons next to each other, a given distance away from similar positive charges, they'll develop the same voltage pretty much independently of how many there are. Increasing the area of a capacitor's plates gives charge carriers more room to spread out -- and, hence, more charge can be stored per voltage, and the capacitance goes up.

* This may just spawn the next layer down of "why" -- if you feel the need to go there, this is treated very well in innumerable physics courses and probably videos as well. Asking the next question down is the start of "please teach me all of electrostatics" -- and that's a course of study that takes years if it's properly backed by mathematics. So if you're interested, consider physics or electronics for your college courses, or your course of self-study.

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Voltage times capacitance is charge stored in the capacitor. Q=C×U.

And since Q=I×t, it takes longer to charge if current is equal.

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    \$\begingroup\$ But on an intuitive physical level, why is that relation the case? \$\endgroup\$
    – ludicrous
    Sep 17 at 17:48
  • \$\begingroup\$ In a simulator, there is none, just the capacitance set by user. Physically, it's only about area and distance as you say. And also about the dielectric constant of the insulating material. \$\endgroup\$
    – Justme
    Sep 17 at 17:50
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Capacitance is charge per volt. More capacitance means you need to supply more charge to change the voltage. Supplying more takes longer.

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The bigger the capacitor, the more charge it takes to charge it up to a given voltage.

The resistors limit the current that can flow in the circuit, so a bigger capacitor will take longer.

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