Capacitor Charge Flow: Examining Circuit Behavior in Multisim

[Mr.Bomzh]

Can a capacitor be discharged/recharged by using only one plate?

I have a theoretical circuit with a transformer winding and a capacitor in series , one end of the winding is at +12v, then the other is connected to the capacitor , and the capacitor other plate connected to ground , what would happen
if I put a switch between the capacitor plate and ground and another swich from the +12v to the ground plate of the capacitor , and then switch them one by one ?

I am trying out Multisim , and I saw this circuit somewhere in internet and i wanted to see how it works in simulation.
Multisim shows me a square wave , when i flip the switches on and off and then off and on.So I wondered what happens in the capacitor when the negative charge is replaced by the positive , does the other positive repels and flows back to the transformer then?

 

[Simon Bridge]

Treat the ground rail as just another wire.

 

[sophiecentaur]

Personally, I would avoid using Simulations to try to resolve the more fundamental questions - such as the one you're asking. Presumably, you do not know the basics on which Multisim is working - or you would not be asking this question. Hence, any answers that Multisim gives you will only tell you want you can expect to measure if you build a circuit. I'm basically saying that your question is at a deeper level than Multisim can answer.
A capacitor 'works' because a PD applied across the plates will produce a big imbalance of charge. A relatively small PD can cause a large separation of charge because the plates have a big area and a small spacing between them. This property is called Capacitance and it can be further increased by putting an insulating 'dielectric' between the plates.
Q = CV
where Q is the charge, C is the capacitance and V is the PD.
There will be the same total number of protons and electrons in an electric circuit so the whole thing cannot be regarded as being 'charged up'. It will be neutral.

If you take an isolated object, you can add or subtract a few (billions? of) electrons. These electrons need to come from (or put) somewhere, which would be the Earth or even at infinity. You need a vast voltage (PD with respect to the Earth or infinity) in order to produce a small imbalance this way so an isolated object has a very small 'self' capacitance. Even the whole Earth has a capacitance of only about 700μF. If you take one wire of a 1000μF capacitor out of the box and try to 'charge it, you will have only the capacitance to Earth (self capacitance) of the metal, which will be only a very few pF. This 'parasitic' capacitance can be a real problem to designers of RF circuits but that's another issue.

 

[Mr.Bomzh]

Ok, I see, it works like this, if I have a metal rod or a single plate, and if I would like to charge it I would have to create a charge imbalance in it and since like charges repel that imbalance would be really hard to make and require high voltage , because voltage is proportional to electric field strength is this correct reasoning?
On the other hand having two rods or plates each with opposite charge draws those carge together and it is then easier to add more charge.
After I have charged a capacitor with PD across it is there any possibility for me to discharge the capacitor or make the charge flow back into the circuit without shorting the capacitor with resistor or other stuff?

 

[sophiecentaur]

Think in terms of the PD needed to make the charges go where you want them to go.
Consider the 'other' plate of the capacitor. The Potential on the charged plate next to it will be V and the Potential of the Earth, many metres away, is Zero. There will be a tiny flow of charge across this second plate. The plate will reach equilibrium when the slight excess charge on the side near the charged plate is equal and opposite to the excess charge on the 'earth' side'. A tiny charge and massive capacity will mean that the Potential Difference across the two plates will be tiny (V = Q/C). The same, tiny charge but really tiny capacitance to Earth will give a high PD to Earth - very nearly the same as for the charged plate. Hence the capacitor will be 'storing' virtually nothing, in the conventional way that capacitors 'store' charge / energy.

This stuff gives many people a headache so don't feel bad if you need to go over it a few times. To be honest, most people don't actually bother except when they're in 'Christmas Puzzle' mood.

 

[Mr.Bomzh]

So let me express the way I understood. Capacitance is proportional to surface are of plates and inversely proportional to distance between them.
So a tiny capacitance is most likely two plates with very samll surface area so if those plates have the same charge as the ones with larger surface area it means that the charge is more squeezed together on them so that forms higher voltage right?

Would it be possible to create a capacitor with varyable dielectric permittivity? A capacitor whose capacitance and hence charge could be controlled?

This is not giving me headache ,although not all of this is easy to understand.
I do have this science explorer type of mood in me, so I hope I will manage.

 

[Simon Bridge]

Variable capacitors exist - they are usually varied by changing the surface area or the distance between the plates. The station tuner knobs on old radios use the variable area kind and electronic balances (used to?) use the variable separation kind.

 

[sophiecentaur]

So let me express the way I understood. Capacitance is proportional to surface are of plates and inversely proportional to distance between them.
So a tiny capacitance is most likely two plates with very samll surface area so if those plates have the same charge as the ones with larger surface area it means that the charge is more squeezed together on them so that forms higher voltage right?

Would it be possible to create a capacitor with varyable dielectric permittivity? A capacitor whose capacitance and hence charge could be controlled?

This is not giving me headache ,although not all of this is easy to understand.
I do have this science explorer type of mood in me, so I hope I will manage.

In a variable capacitor ( moving dielectric or moving plates - it's all the same) the charge stays the same, it's just the PD that will change as you vary the capacity. This, btw, also will involve some (mechanical)energy in or out of the system.

 

[Bobbywhy]

There also are variable capacitors that have no moving parts:
“In electronics, a varicap diode, varactor diode, variable capacitance diode, variable reactance diode or tuning diode is a type of diode whose capacitance varies as a function of the voltage applied across its terminals.”
http://en.wikipedia.org/wiki/Varicap

 

[sophiecentaur]

There also are variable capacitors that have no moving parts:
“In electronics, a varicap diode, varactor diode, variable capacitance diode, variable reactance diode or tuning diode is a type of diode whose capacitance varies as a function of the voltage applied across its terminals.”
http://en.wikipedia.org/wiki/Varicap

... and changing their capacitance involved an energy transfer, too.

 

[Mr.Bomzh]

About the dielectric we are talking about permittivity here right?
The higher permittivity the more charge can be on plates with the same surface area so a better material in a capacitor ?

I know mechanical ones , like in older radios and stuff, I was thinking is it possible to make or maybe there already are , capacitors which dielectric material permittivity could be changed during operation?
I seems as varying permittivity of the dielectric changes the strength of the electric field in that material which changes the strength with which the charges can attract on the opposite plates is that right?

Also would it work like this that having a capacitor with high permittivity dielectric , then charge the capacitor to some value , and then , for example, decrease the permittivity to almoust zero or to vacuum permittivity as it is probabley the lowest , what would happen?
Would most of the charge go away from the plates due to the field dropping down in strength?

 

[sophiecentaur]

The charge remains the same. Where could it go, if it were to be different?
What happens is that the PD changes. If you reduce the capacity by separating the plates, you have to do work on pulling them apart. If you let them pull themselves closer together then you can get work out. Whatever fancy way you can think of for changing the capacity, as I wrote before, energy needs to be transferred one way or the other. You could squirt a dielectric liquid in between the plates to increase capacitance (it would actually be 'sucked in' and you'd have to pump it out to decrease capacitance).
I say again: The Charge Stays The Same!

 

[Mr.Bomzh]

I just thought that if the field gets weaker the charges won't attract as much and since on a single plate we have alike charges they should repell themselves more now because the attracting force has decreased.
So it seems that once you charge a capacitor , no charge will flow to it or away from it , all you can do is change the PD aka the voltage that you would read off from the plates of the capacitor putting a voltmeter across them is that right?

 

[sophiecentaur]

You need to think of it the other way round; somethings can change and some things can't.
What do you know about the field? You are making assumptions and, in any case, the field doesn't actually matter - it's the Potential that counts.
Once you have put the charges on the plates, how did you think they would go anywhere? The whole thing's insulated. What does happen is that, as I said, you have to do work to separate the plates: that is changing the Potential (Mechanical work input gives an increase in the electrical potential)

Think of a mechanical analogue. You separate a mass from the Earth's surface. This doesn't alter the Mass - it just alters the Potential Energy. If you take a mass that is already on a stiff spring and try to replace that spring with a weak spring, the mass stays the same but its position will change and the energy situation will have changed.

[Edit: Yes - a voltmeter can measure the change in volts as you move plates about]

 

[Mr.Bomzh]

Yes I see , that a stiffer spring and the same mass would mean that the mass would stand higher and have larger potential energy than the same mass on a smaller/weaker spring.

Maybe I didn't explain myself , all the time I was kinda reffering to the capacitor we are talking about as connected to a circuit I talked about i the first post.
What happens then , do the charges still stay on the plates constantly even though the permittivity of the dielectric would be varied?

Maybe rhetorically but what does anyone know about the field? Just apart from that it exists and some certain constants about it?...

 

[sophiecentaur]

It depends upon the circuit. If you have a battery connected across the capacitor then, of course, the voltage cannot change and Charge will flow in or out so that things will settle down so that Q = CV again.
The thing about Field is that Field is the Gradient or 'slope' of potential (The Volts per Metre). As you increase the voltage across a fixed capacitor then the Field will go up pro-rata. If you keep the volts constant, then the Volts per Metre will drop as you separate the plates. All that stuff applies but only when you keep the appropriate things constant. A capacitor out on its own (which is what I thought you were talking about) is not so simple to discuss but for an ideal capacitor, doubling the spacing will halve the capacitance, which will double the voltage (V = Q/C). The voltage has doubled and so has the distance, so the field (Volts per meter) is the same as before. Basically, stretching the plates apart has doubled the stored energy. That's one to go away and think about!

 

[Mr.Bomzh]

Ok I understand the plate stretch , voltage increase case.
I will describe a circuit i built in Multisim to ask whether in that circuit the charge flow could be achieved with variable permittivity ok?

It's just a transformer with center tapped primary , the center tap is at +ve, then each side of the primary I attached to a capacitor , the other side of each cap is attached to ground.
What would happen if the dielectric permittivity would be increased /decreased in an alternating fashion? I would like to understand this from a circuit , charge flow point of view.

 

[davenn]

....., the center tap is at +ve, then each side of the primary I attached to a capacitor , the other side of each cap is attached to ground.
What would happen if the dielectric permittivity would be increased /decreased in an alternating fashion? I would like to understand this from a circuit , charge flow point of view.

trying to make sense of what you are saying
if its a transformer then the centre tap won't be at a steady +V

you do understand that transformers work with AC NOT DC ?

in a transformer when connected to an AC voltage on the primary, an AC voltage is generated on the secondary windings.
The voltage is cycling between positive and negative so any capacitor across a set of windings is alternately going to see a cycling + and - voltage on each of its terminals

Dave

 

[Mr.Bomzh]

Exactly they don't work in dc , but they work on anything that changes amplitude or/and polarity.
So I thought about the capacitor and the permittivity of the dielectric and how varying that could cause charge flow back and forth but the only variable capacitors i can find vary the plate surface area contact.

 

[davenn]

Sophiecentaur said it a couple of times earlier
maybe you still missed it

changing the dielectric or the distance between the plates WONT change the charge on the capacitor it will only affect the value of the capacitance

whatever voltage you charge the capacitor to will stay there. It will only change if the capacitor is discharged via a circuit path
1) physical wired circuit
2) leakage between plates across the dielectriccheers
Dave

 

[sophiecentaur]

trying to make sense of what you are saying
if its a transformer then the centre tap won't be at a steady +V

you do understand that transformers work with AC NOT DC ?

in a transformer when connected to an AC voltage on the primary, an AC voltage is generated on the secondary windings.
The voltage is cycling between positive and negative so any capacitor across a set of windings is alternately going to see a cycling + and - voltage on each of its terminals

Dave

I think I know what he's saying. If you change the capacity by altering the dielectric, the Capacitance will change. If you change it in a regular alternating way, then the C will change that way. This will always involve varying the energy situation. Either there will be more or less energy stored in the capacitor or energy will flow out off and into the Power Supply.

When we get to a stage like this, I always suggest that the questioner stops and questions himself again. Is it more likely that he is wrong or that generations of Electrical Engineers are wrong?
Mr.Bomzh. You can either come to terms with this and try to see it in a way that agrees with the established wisdom OR you have to convince yourself that you are the first person in history to see it the 'right way round'.
Is your name Newton, Einstein, Hawkin? You have to make an effort to accommodate, I think.
Some private effort is needed sometimes. You can't expect someone else to flick a magic switch to help you every time. We have done our best.

 

[Mr.Bomzh]

It's ok thank you davenn and sophie you have helped enough and I appreciate that.
I'm not saying I'm right on this one but I'm pretty sure Newton and other shave gotten things wrong in their life also.It's just that some people have the capacity to get them right once in a while while others don't have that.

"I think I know what he's saying. If you change the capacity by altering the dielectric, the Capacitance will change. If you change it in a regular alternating way, then the C will change that way. This will always involve varying the energy situation. Either there will be more or less energy stored in the capacitor or energy will flow out off and into the Power Supply."

Yes sophie you got it right what I was trying to say , so is this right or wrong after all?
I mean the case where the capacitor is attached to a circuit , charged up , then it's dielectric permittivity dropped to say as low as possible , so what happens does the charge flow back into the circuit due to much lower field keeping it on the plates attracted or does the charge stay no matter what and nothing changes?
That would be my last question to which I hope to hear a approving yes or no.
Thank you.

 

[sophiecentaur]

. . . . so is this right or wrong after all?
I mean the case where the capacitor is attached to a circuit , charged up , then it's dielectric permittivity dropped to say as low as possible , so what happens does the charge flow back into the circuit due to much lower field keeping it on the plates attracted or does the charge stay no matter what and nothing changes?
That would be my last question to which I hope to hear a approving yes or no.
Thank you.

The charge will flow out of or back into the circuit (involving some energy flow too) until the condition that the charge Q = CV again.
I don't see why you are still having a problem here. You have 'half' accepted it but don't quite feel it in your bones?
Perhaps your final problem may be because you are still trying to explain what happens in terms of Fields and not Potential. I have been ranting on about this in another thread but there is no reason to think that Field is any more of a fundamental idea than Potential. They are just related by the fact that Field is the Gradient of Potential. If you think 'Potential', then the thing becomes more easy to accept.

 

[Mr.Bomzh]

I guess the bones are the last ones to get familiar with this stuff. :)
Ok so speaking without the term field , charging the cap to some potential and charge involves the movement of charges which transfer energy , then as we drop the permittivity the charges go backwards again we transfer energy , maybe not all of the stored energy pre charging up but atleast some of it.
While we are doing so we increase and decrease the potential between the plates , charging up increases the potential and droppind down in permittivity would decrease the potential between the plates because there would be less charge on them is that correct?

 

[sophiecentaur]

Bones aha

If you just follow the basic rules then the answers have to come out. Keep the PD constant with a source of emf and the charge will flow on or off the capacitor, as you change the Capacitance. Keep the Charge constant (open circuit situation) and changing the capacitance will alter the PD. Remember there are two very distinct situations - open circuit and powered circuit - with very different outcomes.
Is there anything else about this to discuss now?

How you choose to alter the Capacitance makes no difference to that rule.

 

[Mr.Bomzh]

Oh thanks , that was just what I thought , thank you for clearing my doubts.I guess for now I have sorted things out , for this one at least.

 

[sophiecentaur]

One satisfied customer. Brill.