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10kv and 30kv capactor plates

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Andy
Tue Dec 17 2013, 11:07PM Print
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi
This might be a basic question, but if one plate of a capacitor was connected to +10kv supply, and the other side connected to a -30kv supply, would the +10kv side increase in voltage to +15kv or +30kv, or stay at +10kv.

Thanks
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Proud Mary
Tue Dec 17 2013, 11:58PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
You are talking about matters of potential difference here - difference related to a something which is normally taken to mean Earth.

Therefore, if you have one plate of your capacitor that is at 10 kV relative to Earth, and the other plate has a voltage of -30 Kv relative to Earth upon it, then the potential difference between them is 40 kV.

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BigBad
Wed Dec 18 2013, 12:00AM
BigBad Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600
A metal surface is an isopotential for DC. so it stays at 10kv.

To achieve that, current necessarily flows in from the power supply to do that though; so if you unplug it, and then charge the other plate it will of course vary.

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Andy
Wed Dec 18 2013, 12:03AM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
@Proud Mary
You didn't really answer the question, or I expanied it wrong...But will the 10kv one rise to +20kv, if its half of 40kv. If I have two transformers with one supplying one plate to 10kv, and I raise the other plate higher and higher, will the electric field of the 10kv plate increase to 20kv or stay at 10kv to earth?

@BigBad
Why would you have to unplug the source? If I had +10kv and -10kv and then raised the -10kv to -30kv, with the transformer(at dc) still connected to the +10kv(not including the return path), your saying that I would have to disconnect the + side transformer for the plate to rise.
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Houdini0118
Wed Dec 18 2013, 01:15AM
Houdini0118 Registered Member #8558 Joined: Thu Dec 06 2012, 11:38PM
Location: usa
Posts: 57
No the capacitor does not have to center around 0v. You can have a capacitor with one side grounded and charge the other side to 100v and it wouldn't make ground 50v.
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BigBad
Wed Dec 18 2013, 01:45AM
BigBad Registered Member #2529 Joined: Thu Dec 10 2009, 02:43AM
Location:
Posts: 600
Andy wrote ...

@BigBad
Why would you have to unplug the source? If I had +10kv and -10kv and then raised the -10kv to -30kv, with the transformer(at dc) still connected to the +10kv(not including the return path), your saying that I would have to disconnect the + side transformer for the plate to rise.
Any electrically connected metal is going to be at exactly the same DC voltage all over it.

The physics and the reason is that if any charge on the plate or the lead was at a different voltage/potential energy, electrons would immediately move to even out the potential, flee from high voltage potential and attracted to low potential until they're all at the same potential all over. Not the same charge density, you understand, the same potential.

Only if the current can't flow fast enough into or through the plate will the voltage vary across the plate; in the case you're describing nothing is changing nearly fast enough for anything to budge noticeably.

So if you have a power supply with +10kv, and you connect a plate to the 10kv terminal, then the power supply lead and the plate will go to 10kv and be held there by the power supply; and you can wave any charged plates of any voltage (within reason) around near that, and the plate will stay at 10kv.

Some parts of the plate may have more charge than others though, but they will all be at exactly the same potential.

But if you were to charge the plate up to 10kv, and then disconnect it from the power supply, and then (say) move or charge a 30kv plate near to it, the potential on that plate would then change, it would be higher- the same everywhere on the plate, but at higher potential, like 15kv or something. If you put a -30kv plate near it, it would go down 7kv or something.

But it would only do that if it was disconnected from the power supply.

(Actually, I'm talking about metal plates in space, which is a slightly more general question, but two pole capacitors work exactly the same way, air is a dielectric, but you can't change the geometry with a two terminal capacitor.)
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Andy
Wed Dec 18 2013, 03:33AM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874

So if you have a power supply with +10kv, and you connect a plate to the 10kv terminal, then the power supply lead and the plate will go to 10kv and be held there by the power supply; and you can wave any charged plates of any voltage (within reason) around near that, and the plate will stay at 10kv.
How high would the charge have to be above reason?


But if you were to charge the plate up to 10kv, and then disconnect it from the power supply, and then (say) move or charge a 30kv plate near to it, the potential on that plate would then change, it would be higher- the same everywhere on the plate, but at higher potential, like 15kv or something. If you put a -30kv plate near it, it would go down 7kv or something.
I don't understand why you would have to remove the supply, but that is what I was thinking about. Isn't 30 kv high than 10kv so more electrons or lack of them will move throught the plate more and raise the voltage, or if minus stop charge going onto the plate.


No the capacitor does not have to center around 0v. You can have a capacitor with one side grounded and charge the other side to 100v and it wouldn't make ground 50v.
But each plate would hold have the voltage wouldn't it, 50% electrons, 50% protons?
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Dr. Slack
Wed Dec 18 2013, 07:11AM
Dr. Slack Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
If you connect a plate to a supply, then you have fixed its voltage.
If you change the electric field around it by changing the voltage of a nearby plate, or the position of a nearby charged plate, then charge will flow into or out of the supply along the connecting wire, to maintain the plate at the same supply voltage.

If you have an isolated plate, then you have fixed its charge.
If you change the electric field round it, then the voltage of the plate will change.
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Proud Mary
Wed Dec 18 2013, 12:45PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
I wonder how it would change things to have number of extended plates of small radius arranged parallel to eachother in the form of insulated wires, and all these wires had different voltages on them as in a multi-way cable, or, in a word, a bunch of wires!

Now let's say we just have two parallel wires in a length of electric flex/cord. One of these two insulated wires is connected to the negative pole of a 30V battery, the other terminal of which is earthed. The second wire is connected to the positive terminal of a 10V battery, the other terminal of which is also earthed to a common ground.

Beyond the momentary pulse of current from the batteries as they charge up the capacitance formed between the two insulated wires, does any current flow? No, it doesn't (except for ~picoamp leakage through the insulation, which is not at issue in the present discussion).

Can anything happen at all if one side of each battery is not connected to Earth/chassis common ground? No.

See how folk are bewitched and lose their senses as soon as kilovolts are mentioned! smile
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Andy
Wed Dec 18 2013, 09:48PM
Andy Registered Member #4266 Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
I think I understand it.
Was planning on building a marx and trying to get the output to feedback into the stages with a transformer supplying the difference. If there is three plates, the middle one +15volt with the left side -15volt the left side is connected in series with the next capacitor and then connected to the right side which is in parallel of the two capacitors.

If that makes sense tongue

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