Question: 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.

[![enter image description here](https://i.stack.imgur.com/IibM6.png)](https://i.stack.imgur.com/IibM6.png)

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

[![enter image description here](https://i.stack.imgur.com/NXrgW.png)](https://i.stack.imgur.com/NXrgW.png)

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?

Answer: 
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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.
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.