What would happen if I take apart the plates of a charged capacitor?

  • #1

Main Question or Discussion Point

what would happen if I take apart the plates of a charged capacitor? Would the plates stay charged, or will the plates loose their charge? will the energy stored in the capacitor during charging stay stored in the plates though there is no more existence of electric field?
 

Answers and Replies

  • #2
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If you disconnect the plates from the battery and then move the plates apart . Then yes the charge will stay put, it has no place to go .
 
  • #3
Drakkith
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what would happen if I take apart the plates of a charged capacitor? Would the plates stay charged, or will the plates loose their charge? will the energy stored in the capacitor during charging stay stored in the plates though there is no more existence of electric field?
The electric field still exists. One plate will be positively charged with respect to neutral, and the other will still be negatively charged. At least until they touch something and discharge.
 
  • #4
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What do you mean by neutral?
 
  • #5
Drakkith
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What do you mean by neutral?
If capacitors work the way i think they work, but adding electrons to one plate and removing them from the other plate, then they are charged with respect to neutral. If you discharge the capacitor completely, then both plates have no charge and are neutral.
 
  • #6
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The charge will remain however the energy will not be the same. There is energy stored in the electric field itself. If move the plates you will be doing work on the system.
 
  • #7
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When you move the plates apart the voltage will increase. Move the plates closer together and the voltage will decrease. That's the way condenser microphones work. Sound waves cause the plates to move closer together and farther apart. The changing voltage is then amplified by the pre-amp.
 
  • #8
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When you move the plates apart the voltage will increase. Move the plates closer together and the voltage will decrease. That's the way condenser microphones work. Sound waves cause the plates to move closer together and farther apart. The changing voltage is then amplified by the pre-amp.
I don't think most mics work like that, usually they work on Faraday's law. I think keyboards use capacitance to record keystrokes.
 
  • #9
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I don't think most mics work like that, usually they work on Faraday's law. I think keyboards use capacitance to record keystrokes.
I did not say "most microphones". I said http://en.wikipedia.org/wiki/Microphone" [Broken].

Edit:
According to the Wikipedia page "the vast majority of microphones made today are electret condenser microphones".
 
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  • #10


will the energy decrease or increase if you disconnect the plates from the battery and then move the plates apart??

can charge be kept in a plate without energy??

if the energy stored in a capacitor is related to the electric field, the how come we will still have stored energy though we move the plates for a long distance??
 
  • #11
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disconnecting the plates and then increasing the distance would increase the energy stored in the fields, because you are creating E fields where they didn't exist and The E field is constant in between the plates. The energy stored in a capacitor is 1/2CV^2
 
  • #12
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Capacitance won't stay constant though. It is equal to the area of the plates divided by the distance between the plates. Capacitance would approach zero in this scenario, wouldn't it? Which means your energy would be decreasing???
 
  • #13
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Capacitance won't stay constant though. It is equal to the area of the plates divided by the distance between the plates. Capacitance would approach zero in this scenario, wouldn't it? Which means your energy would be decreasing???
You are correct that capacitance would decrease, but the plates were charged with the higher capacitance when the plates were close together, so they will no loose charge or energy, assuming they were not permitted to discharge. If you wanted to recharge the plates after they were moved farther apart, at the same voltage, the capacitor would store less energy.

After charging the plates, they want to attract -- opposite charges attract. To pull the plates apart would require energy, and this energy would go into increasing the voltage between the plates. The voltage would be linearly related to the distance at first, but as the plates moved farther and farther apart, the assumption that the field was uniform inside the plates would no longer be valid, so the voltage would increase more slowly.
 
  • #14
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I guess I'm having a hard time wrapping my head around this. pulling the plates apart would really increase the voltage between them? Wow... The energy that you put into the system to pull the plates away from each other will increase the voltage on the plates... Seems like this might be a useful phenomenon in certain situations. Is there a component that utilizes this ability?

I know condensor microphones were mentioned earlier... the way I understand their operation is that when the pressure (sound wave) reaches the plates of the mic, the plates come closer and farther apart, which increases and decreases the voltage accordingly and gives you an output that is an electrical representation of the soundwaves at the input. Is this correct?

Anyway... back to the capacitor... the charges only "built up" on the plates in the first place because they were attracted to each other by an electric field. Now... as the plates seperate, this field becomes weaker and weaker and one would assume that if the plates kept moving farther away from each other, they would escape the influence of the field altogether and they would simply become metal plates... presumably with a charge still on them. We've basically just turned these two plates into a "static" situation where we could move them close to the ground and produce mini lightning bolts, right? Sounds like a neat science project.
 
  • #15


if we assume that these two charged plates are separated for a long distance in a way that the influence of the electric field no more exists, how could we calculate the voltage between these two plates?
 
  • #16
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I guess I'm having a hard time wrapping my head around this. pulling the plates apart would really increase the voltage between them? Wow... The energy that you put into the system to pull the plates away from each other will increase the voltage on the plates... Seems like this might be a useful phenomenon in certain situations. Is there a component that utilizes this ability?
None that I know of -- an interesting idea, though.

Anyway... back to the capacitor... the charges only "built up" on the plates in the first place because they were attracted to each other by an electric field. Now... as the plates seperate, this field becomes weaker and weaker and one would assume that if the plates kept moving farther away from each other, they would escape the influence of the field altogether and they would simply become metal plates... presumably with a charge still on them.
Well, the electric field would get weaker, and I suppose at some point it would be negligibly small, but it won't disappear until the charge does. (Unless there's another conductor to interact with.)

We've basically just turned these two plates into a "static" situation where we could move them close to the ground and produce mini lightning bolts, right? Sounds like a neat science project.
Yeah, we're dealing with static electricity and if the voltage were high enough, you would get a noticeable arc.

brainyman89 said:
if we assume that these two charged plates are separated for a long distance in a way that the influence of the electric field no more exists, how could we calculate the voltage between these two plates?
There are two easy models for dealing with capacitor plates. When the plates are very near each other, we can assume the field inside is uniform (which is only actually true when the plates are infinite planes), and when the plates are far away, we can look at them as point charges (which is only approaches reality as they get farther away).

Here's a simple formula for voltage due to point charges.

[URL]http://maxwell.byu.edu/~spencerr/websumm122/img213.gif[/URL]

where r is the distance, q the charge, and k a constant. OK, but this is a bad way to write this formula for our purposes, though, since we know V is maximum at infinity, then decreases as the plates move closer. We really want V = C - kq/r (we can add the constant both because voltage is relative and because this formula comes from an integration problem), where C is a constant that can be found if you know the voltage of the plates when they are at some distance r.

At distances where the infinite plate or point charge methods don't work well enough, the math would get more complicated; the charge distribution would no longer be uniform on the plates. To be honest, calculating voltage in that region is a little beyond my skill.
 
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