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: 
[There are two closely related notions of capacitance: self capacitance and mutual capacitance](https://en.wikipedia.org/wiki/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](https://i.stack.imgur.com/mO6cr.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fmO6cr.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)
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.