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Imagine a battery that creates a +9000V Potential difference in respect to ground (The Earth) in it's positive terminal. (The Earth is at 0V in this example)

And a +8992V potential difference in respect to ground in it's negative terminal.

  • Suppose we connect this battery to a capacitor with a voltage rating of 8V, Normally it shouldn't exceed the capacitor voltage rating due to 8V potential difference.

First Question: Will this capacitor have a higher capacitance(even a tiny bit more capacitance?) Second Question: In a closed circuit this capacitor will produce only a potential difference of 8V, But if we connect the positive terminal to ground,And the negative terminal to ground, Will the potential difference be much higher?.

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With the way you constructed it, the capacitor will have precisely the capacitance it was measured to have. Capacitance is based on current and voltage difference across its leads. It matters not where the "ground" reference is.

You will see something called a "parasitic capacitance" because of the large ground plate beneath you. However, this has nothing to do with the capacitor itself in any way. Its the geometry and the atmosphere that matter there. If you took the capacitor away, the parasitic capacitance with the ground would remain.

If you connect the leads to ground, that will change the voltages accordingly. You'll probably get some arcing, because you have devices built for 8V that suddenly have 9kV across them. The exact behavior of how that voltage changes is a more complex transient question. But in the end, you'll have a new voltage across the leads of the capacitor, and it will charge accordingly.

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  • $\begingroup$ Thanks, But when the capacitor is going to get charged at +9000v shouldn't it just hold a bit more charge so it reaches +9000v (like a tiny bit more)? $\endgroup$ – mohamed azaiez Oct 29 '19 at 10:50
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    $\begingroup$ @mohamedazaiez I'm not 100% sure what you mean, but if I venture a guess, the entire capacitor will be given a static charge of an average of 8996V, so it will have fewer electrons on it than a netural capacitor would. There would be a electrostatic attraction trying to pull the capacitor towards the earth. However, when we talk about measuring a capacitor's charge, we are talking about the potential between the plates, which is a difference between them. The difference is only 8V, no matter where you put the system, because that's the voltage difference your battery provides. $\endgroup$ – Cort Ammon Oct 29 '19 at 15:48
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    $\begingroup$ This is actually an issue for electronics, called Electrostatic Discharge. It is possible for a system to build up a static charge of several thousand volts without the operation of the system being affected -- all the capacitors just work like normal. At least, they work like normal until someone touches them, creating a path for that high static voltage to flow. It burns some parts out (usually we phrase that as the part staying at "ground" voltage, and our bodies being charged with a static charge, but from an E&M perspective, they're identical) $\endgroup$ – Cort Ammon Oct 29 '19 at 15:51

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