Capacitor Q

[andafoo]

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When the potential on a capacitor is removed or turned off will the charge remains or will it discharge?

For some reason, seems like I've seen both discharging happen as well as it remaining charged. Last time I saw an actual capacitor was 3-4 years ago... so I can't really remember what happens!

 

[fj4288]

I'm guessing it discharges because Q=CV and if you remove the potential, i.e. V, then 0 multiplied by C= 0 Q. I'm not sure if it's right someone please chime in!

 

[Tutmos]

Does the capacitor still have a route to ground or are you pulling it out of the system? If you're pulling it out of the system or "throwing a switch off" then the charge will remain in place. If turning the voltage down toward zero then yes it will discharge.

I voided today, take my opinion with a grain of salt

 

[andafoo]

Does the capacitor still have a route to ground or are you pulling it out of the system? If you're pulling it out of the system or "throwing a switch off" then the charge will remain in place. If turning the voltage down toward zero then yes it will discharge.

I voided today, take my opinion with a grain of salt

You sound like you know it well though 😀. It makes sense that if you open the circuit the charge remains.

But what is the difference between turning your voltage to 0 and opening the circuit... just the physical path for charge to travel?

 

[andafoo]

I think by "turning the voltage to 0", you mean removing the battery that was previously holding the potential. If so, by removing the battery, and if the whole circuit is still connected, the capacitor will discharge.

You may have to describe it in further detail. If you remove the battery, how is the whole circuit still connected?

I may also need to back up some. When a capacitor is fully charged... the current will be 0? That means we have a buildup of charge on one of the plates ---- and ++++ on the other.

To discharge it, would that mean that the electrons (----) need some way to distribute back, lowering the potential?

Can someone explain this better?

 

[Kaustikos]

You may have to describe it in further detail. If you remove the battery, how is the whole circuit still connected?

I may also need to back up some. When a capacitor is fully charged... the current will be 0? That means we have a buildup of charge on one of the plates ---- and ++++ on the other.

To discharge it, would that mean that the electrons (----) need some way to distribute back, lowering the potential?

Can someone explain this better?

The easiest way for me to understand capacitors is to think of it as a reservoir and the circuit as a verticalish pipe system of water. Hook up the capacitor and pour water (The batter) down the pipe. The capacitor will catch some of the water and "fill up" with water. The capacitor/reservoir is charged/filled up with water and then the rest continues down. If you stop running water (battery removed/turns off), then the capacitor/reservoir will use up whatever charge/water it has and continue the circuit, discharging itself.

 

[spooge]

The potential remains until you complete the circuit...you always have to check to be sure your capacitor has been discharged before you put it on your board or use it in your experiment(aka you complete the circuit by joining the two plates). The potential does not last forever but if you dont believe me run the current through the capacitor for thirty seconds then remove the source. Wait five seconds and stick your tongue in between...theory and applications are different...which is why you have physicists and engineers.

 

[andafoo]

The potential remains until you complete the circuit...you always have to check to be sure your capacitor has been discharged before you put it on your board or use it in your experiment(aka you complete the circuit by joining the two plates). The potential does not last forever but if you dont believe me run the current through the capacitor for thirty seconds then remove the source. Wait five seconds and stick your tongue in between...theory and applications are different...which is why you have physicists and engineers.

Heh, so your saying even if you don't have any physical link between the two plates the capacitor will still discharge? So in general, a capcitor will never hold its charge, so by removing the battery or opening the circuit via switch, the capacitor will still discharge?

 

[spooge]

In a time independent refrence frame the potential lasts forever (weird contrast) and discharge only occurs either after the potential is great enough to complete the circuit or if you connect the plates...we would always connect the plates with a wire and alligator clips on both ends connecting the two plates to make sure it had completely discharged...In reality the potential remains for a while after being removed from the source...The problem I remember having with the MCAT is when they would talk about situations like this and I was not sure if I should be a physicist and answer the question based on theory or be an engineer and answer the question based on reality...if they gave me a graph then it was kosher...I just go by their results regardless of what I know...if it was calculatory then I just used Ohm and the other laws of circuits

 

[kish180]

if you simply remove the battery but connect the circuit, the current will flow, in the opposite direction which the battery provided, because of the capacitor (which is discharging). however since the cap. is simply a reserve of power (or electrons), as it discharges it gives up electrons, reducing its potential and thus reducing the current. The current asymptotically comes to a stop, much like how it behaves when the capacitor is charging. If you dont connect the circuit after you remove the battery, but you just leave the capacitor there, it wont discharge immediately but it may experience dielectric breakdown in which the electrons essentially "jump" between the two plates, taking a path through the dielectric material (which can be air) in order to discharge the potential. This is the phenomenon which results in lightning. This would NOT, however, occur in a vacuum, but it would occur even in a vacuum if you shined light above a threshhold frequency onto the anode. This is a specific application of the photoelectric effect and is something ive seen twice on AAMC passages so id be familiar with it.
Also, spooge you mentioned "In reality the potential remains for a while after being removed from the source" - in theory the potential doesnt just remain for a while, it remains forever - the potential as well as the current declines asymptotically during discharge and charging of the cap. we say that the cap. is completely discharged however when the current it produces is "close enough" to 0.

 

[AWhitehair]

Very simple question...why is it that the current stops when the capacitor is charging? Isn't the current needed to charge the capacitor?? I really wish I would have paid more attention in Physics II lab!

if you simply remove the battery but connect the circuit, the current will flow, in the opposite direction which the battery provided, because of the capacitor (which is discharging). however since the cap. is simply a reserve of power (or electrons), as it discharges it gives up electrons, reducing its potential and thus reducing the current. The current asymptotically comes to a stop, much like how it behaves when the capacitor is charging. If you dont connect the circuit after you remove the battery, but you just leave the capacitor there, it wont discharge immediately but it may experience dielectric breakdown in which the electrons essentially "jump" between the two plates, taking a path through the dielectric material (which can be air) in order to discharge the potential. This is the phenomenon which results in lightning. This would NOT, however, occur in a vacuum, but it would occur even in a vacuum if you shined light above a threshhold frequency onto the anode. This is a specific application of the photoelectric effect and is something ive seen twice on AAMC passages so id be familiar with it.
Also, spooge you mentioned "In reality the potential remains for a while after being removed from the source" - in theory the potential doesnt just remain for a while, it remains forever - the potential as well as the current declines asymptotically during discharge and charging of the cap. we say that the cap. is completely discharged however when the current it produces is "close enough" to 0.

 

[AWhitehair]

Nevermind. This website cleared everything up: http://www.tpub.com/neets/book2/3b.htm