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I just started studying electrical engineering. So far I'm trying to figure things out by visualizing the electron flow and density in different scenarios. But I'm having a hard time when it comes to this particular case.

For example, in this configuration the capacitor is fully charged, and the voltage difference between ground and the lower pin on the capacitor is essentially 0.

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When the switch is opened , the voltage difference between ground and the same pin is suddenly 5 volts.

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I understand that ground should be seen as a zero level for the circuit, and following that rule this makes sense. I just can't quite grasp how this works on the electron level. I believe the capacitor pin that's connected to ground has a higher density of electrons than ground itself, but because the other side of the capacitor has a lower density of electrons, it ends up "pulling" the electrons on the other side. And that's why the potential difference between ground and the lower pin ends up as 0.

Is this correct? If we removed the plate on the opposite side of the ground pin, would there be a potential difference between ground and the plate that's left?

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3 Answers 3

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So far I'm trying to figure things out by visualizing the electron flow and density in different scenarios.

I believe this is a tempting but often misleading way to learn; electrons themselves do lots of counter-intuitive things, and for most circuit analysis problems you're better off thinking simply in terms of "charge".

It is also important to remember that "ground" is just a label and that every component has only "local knowledge".

So, in the classical charge-and-voltage-fields model: the capacitor gets a charge difference such that the potential difference across its plates is 5V, and the top plate is the more positive. (This will correspond to more electrons on the negative plate).

When the switch is thrown, there is a loop allowing current to flow around from the top plate of the capacitor, through the resistor, and back to the bottom plate of the capacitor. It does so.

Looking from the ground, it is now connected to the top plate of the capacitor. The top plate is 5V more positive than the bottom plate, so we label the bottom plate -5V. This will decrease as it discharges.

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  • \$\begingroup\$ Thanks for the reply. So is it correct to think that the plate connected to ground has a higher charge than ground and the plate on the opposite side has a lower charge than ground? \$\endgroup\$
    – Kimple
    Sep 10, 2020 at 9:36
  • \$\begingroup\$ Cautiously I'd say yes; the important thing is that the -ve plate will be higher net negative charge than the +ve plate. But I think that reasoning relative to "ground" gets hard to keep track of once you have multiple capacitors and is likely to mislead you very quickly, e.g. in things like the Cockroft-Walton multiplier. \$\endgroup\$
    – pjc50
    Sep 10, 2020 at 9:40
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There are two closely related notions of capacitance: self capacitance and mutual capacitance. The capacitor's mutual capacitance arises because its two plates, separated by a dielectric material, can store electrical charges, each plate store charges of opposite signs. Besides, the capacitor, as any other body, can store an excessive "common" charge of either sign, and this capability defines the capacitor's self-capacitance. In a way, self-capacitance is a mutual capacitance of a compound capacitor having two plates, one is the capacitor body, the other is a common "ground".

In your circuit, the 200uF capacitor has a mutual capacitance of 200uf, and an unspecified self-capacitance. When you flip a switch, the charges redistribute between the capacitor body (including its plates and dielectric) and the ground, while the mutual capacitance provides the voltage drop that starts the current through the resistor. When the charges stored due to self-capacitance redistribute through the circuit, the potential at the node between the capacitor and the resistor changes from zero to minus five volts. The details of this redistribution also depend on configuration of wires and the resistor, but these factors balance each other in such a way that the voltage drop across the resistor equals the value of the capacitor voltage, because the capacitor's mutual capacitance is much greater than its self-capacitance.

schematic

simulate this circuit – Schematic created using CircuitLab

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If the capacitor is charged to 5V over its terminals and you flip the switch, it will still have 5V over its terminals and it starts discharging. What might confuse you is that the switch sets the absolute voltage of the capacitor positive terminal to either 5V or 0V, which will also change the absolute voltage of capacitor negative terminal from 0V to -5V as the capacitor has 5V over it. Try swapping the resistor and capacitor if it makes more sense to you.

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  • \$\begingroup\$ Thanks, that clarifies why the -5V suddenly appeared. But did I understand correctly, the electron density in the negative terminal is higher than that in ground? Or should I just skip worrying about ground entirely since it's not part of a closed system? \$\endgroup\$
    – Kimple
    Sep 10, 2020 at 9:14
  • \$\begingroup\$ The density on the -ve plate will be higher than the -ve plate, but how this relates to the ground label isn't especially relevant. \$\endgroup\$
    – pjc50
    Sep 10, 2020 at 9:28

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