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