Understanding Capacitor Voltage in Relation to Ground

[Mzzed]

If a capacitor is charged to 10V where the negative side is connected to ground (0V), when the capacitor is disconnected from the power supply on both the positive and negative sides;

1) Will the negative side of the capacitor still be 0V relative to the ground it was just connected to?

2) Say the two sides of the capacitor are shorted. Charge would flow from the positive to the negative side of the capacitor, so does this mean the negative side of the capacitor will no longer be the same voltage as the ground it was connected to previously?

I'm asking this because I have components connected to the negative side of a capacitor that will likely break if the voltage rises too much above ground and although I think I know the answer to this I am really doubting myself now.

Any help is appreciated, thanks!

 

[davenn]

I'm asking this because I have components connected to the negative side of a capacitor that will likely break if the voltage rises too much above ground and although I think I know the answer to this I am really doubting myself now.

what other components ? show us the circuit

 

[Mzzed]

I was just generalising, the actual circuit is higher voltage and I don't have a diagram of it sorry. But for example I know the negative of a 1.5v battery is attached to the negative pin of the capacitor.

 

[Rive]

... I have components connected to the negative side of a capacitor that will likely break if the voltage rises too much above ground

If those components are lifted together with the capacitor and has no connection to the ground then they will remain on the voltage of that capacitor pin where they are connected, so they won't break down.

If those components are connected to the ground and the capacitor too, then the capacitor is not actually 'disconnected' from ground: it still connects through those components -> more details needed.

 

[Mzzed]

If those components are lifted together with the capacitor and has no connection to the ground then they will remain on the voltage of that capacitor pin where they are connected, so they won't break down.

If those components are connected to the ground and the capacitor too, then the capacitor is not actually 'disconnected' from ground: it still connects through those components -> more details needed.

ah ok, well despite that what happens in the situations 1) and 2) I mentioned before, ignoring any other components?

 

[anorlunda]

The absolute potential of a node depends on the rest of the circuit between the node and ground.

In circuits, we calculate voltage differences between two points. The absolute potential and the choice of where to connect ground ( if at all) is rarely significant.

If you are trying to understand circuits and ideal components, you're better off forgetting about ground and absolute potentials. Just remember that a voltmeter has two leads, not one.

 

[jim hardy]

ah ok, well despite that what happens in the situations 1) and 2) I mentioned before, ignoring any other components?

I was just generalising, the actual circuit is higher voltage and I don't have a diagram of it sorry.

You do realize you're asking people to guess at what you have in mind. Even in "Charades" you have to give better hints than that.

Do you have a voltmeter and a ladder ? Try it out. One experiment is worth a thousand expert opinions.

 

[Windadct]

You are asking a theoretical question - about a real world situation? "the actual circuit is higher voltage"

I think if you sketch the scenario, and look at how you are going to charge a capacitor - you will find your answer.

 

[Mzzed]

Thanks guys, yeah my bad, should have included diagrams. I'll try testing myself and see if it makes sense.