Is the electric field stronger in resistors with higher resistance?

[Mr.Bomzh]

Hi, I was reading a paper on circuits, electric field and resistance.Now i have a question , the paper says that in a resistor or part of circuit which has higher resistivity the electric field is higher than
in those parts were there is low or lower resistance, now is that true, it seems to me.
So here is my attempted explanation, for a given PD , for example 100 volts , there is a given strength of the electric field as that corresponds to the voltage or PD.
If the wire leading to the resistor has a resisance of 1ohm then the electric field has to do very little work to get the charges moving , in the resistor , for example 100ohm, the electric field has to increase because now it is 100x times harder to push the charges through that part of the circuit, so
the electric field outside and around the resistor is stronger?
like in the water flow analogy for a given water flow , if the pipe gets smaller in diameter the pressure increases.

 

[Simon Bridge]

The electric field is proportional to the gradient of the potential.

 

[sophiecentaur]

If you are really desperate to use Electric Field in the description of the way a circuit functions you need to consider each incremental section of the circuit in detail and it is unlikely to get you anywhere. The more fruitful way to look at it is in terms of Potential and Potential difference (which is how it's done, of course). I have made this point many times but there is nothing more or less fundamental about Potential or Field so using Fields is not, in fact, introducing anything deeper of more meaningful. I know that Forces can seem easier, intuitively, but they may not actually help in the long run. Objects move towards regions of lower potential and the actual topographical route a charge takes (at DC) is often irrelevant to the functioning of a circuit. Ask yourself - does Ohm's Law mention Field?

As for your argument involving resistors, if you make a resistor of the same value, twice as long (there are many different lengths of 100 Ohm resistor available in the catalogue) then the field will be twice as much for the same volts and current. 'Field' has let you down there, hasn't it? This is not to say that, in the particular context, that paper was actually wrong. You may just reading something into what was written that was not intended.

Stick to PD (Volts) and, if you don't find them friendly at this point, learn to use them. They will 'grow on you'.

 

[Mr.Bomzh]

I actually find PD very unfriendly , he has been very rude and abusive to me several times , let's say our interference was quite shocking.. :D

Ok hmm wait , about the 100ohm resistor , so what if I make a 100ohm resistor a 1km long? does the field just keeps increasing , but how high does it increases , obviously for a given PD say 1000v the resistance it can overcome is not infinite , so if we increase the resistance after one point current will stop to flow entirely.what happens then to the field?
obviously stretching the conductor or resistor but keeping the resisatnce the same is different , ok what would happen if I would managed to make a 100ohm conductor so long that it makes a circle around the globe?
I'm asking this because , yes I know you don't like when amateurs speak about fields but let's give a try, Well the electric field strength is more commonly told in the very cute and charming PD or in even more common everyday language volts.I also know that the electric field strength is measured in volts per metre , and it falls of just like gravity with distance and the relationship I read is 1r2. The inverse square law I guess ,

oh just a quesion does that imply that if I have a point charge somewhere in space and when I get half distance away from it the field has dropped 4 times in strength as compared to that at the point charge?

Now I ask the question about the same resistivity but longer resistor because I've read that the electric field is the one that exerts force on charged particles to move , so a given PD has a given strength of E field then how come the e field be twice as strong keeping everything the same just increasing the distance ?
The only thing that comes in mind is conservation of energy, If I have 100 volts , and a wire with a resistor in the middle which is say 100ohms , then from ohm's law I can say to get to the current that if V=IxR then I must be I=V/R so I =100/100 = 1amp of current.
Now that one amp of current must flow in all cases if the resistance is kept the same just the length of the resistor is increased so that energy must be conserved right? Is that the reason the electric field must get up to maintain the same current flow since the resistance has also stayed at the same value?

Another thing I would like to ask is about the E field as said volts per metre.That would imply that if I have a battery which has a PD of 100 volts , then 90cm away from the battery the field gradient should be what 10v?
Now here comes the part I wonder , if the field gradient decreases with distance then what happens if I attach a long cable to the positive terminal of the battery , say the cable is 100m long , the battery still has only 100v PD , is the 100v per metre now measured from the battery or from every place where the extended cable goes and as far as it goes?

 

[ZapperZ]

I actually find PD very unfriendly , he has been very rude and abusive to me several times , let's say our interference was quite shocking.. :D

Ok hmm wait , about the 100ohm resistor , so what if I make a 100ohm resistor a 1km long? does the field just keeps increasing , but how high does it increases , obviously for a given PD say 1000v the resistance it can overcome is not infinite , so if we increase the resistance after one point current will stop to flow entirely.what happens then to the field?

You may not believe it, but you already know about potential difference quite a bit, if you've already taken basic kinematics.

Look at the picture below of two inclined planes of equal height.

Now, what do you think is the difference in potential energy when an object moves, say, from the top of the inclined plane to the bottom? Does the fact that one has a longer ramp than the other makes a difference at all?

Go back to your question. If you have 100 Ohm resistor that is either 1 cm, or 100 meters, connected to the same potential difference, do you think there will be a difference?

Zz.

 

[sophiecentaur]

@ ZapperZ
You beat me to it. I was composing a long rambling explanation but you have said it all. If you needed to drive a car up those two slopes (aka same PD) as fast as you could (max power), you would just use different gears but would arrive at the top in more or less the same time. The force from the wheels (aka Field) would be different in each case but that would be irrelevant.

I do need to correct that piece of idiocy in my last post. Of course, increasing the length of the resistor Decreases the field (the Volts per Metre). I should read through things more carefully before pressing the go button.

 

[ZapperZ]

Thanks.

I've even worked out some of the details in my PF Blog:

https://www.physicsforums.com/blog.php?b=4290

Zz.

 

[sophiecentaur]

A useful bit of writing ZZ.
It's strange that students seem to be quite happy about rolling cars down slopes but the electron volt, as a concept seems to get 'em confused.

 

[Mr.Bomzh]

So a longer resistor of the same resistance doesn't produce a stronger e field in fact the opposite a weaker one if measured in a single point in the resistor compared to that of the shorter resistor, but since the distance is longer the overall field exerted force on charge is the same, because a given PD has only a given amount of force which cannot get bigger by itself right ?It kinda seemed like that, following my inner logic which is not always the best one ofcourse.

Oh and by the way , if I have a circuit with a wire of negligible resistance and a resistor of 100ohm in the middle then the field in the resistor will be larger as compared to that of the wire which had very low resistance?
It kinda seemed like that, following my inner logic which is not always the best one ofcourse.

I guess upon this recent misunderstanding you missed out on the other quesion, about the battery terminals and extended length of wire attached to it and the volts per metre thing?
last paragraph of my previous post. :)

 

[thegreenlaser]

I'm going to assume that you need an electric field interpretation to help understand that paper. However, if you're talking about the electric field, you should be talking about point quantities like resistivity, conductivity or resistance per unit length, rather than bulk quantities like resistance.

In this particular case, resistance per unit length is probably easiest (you could use resistivity, but then you have to worry about whether the cross-sectional area changes). The current in a wire is proportional to the electric field divided by the resistance per unit length. That's a direct consequence of either form of Ohm's law (V = IR or J = E/ρ) and the fact that E = -dV/dx

So if the current is the same and the resistance per unit length is higher, then the electric field must also be higher. Assuming that both have the same current running through them (which they would if they're part of the same loop) a 1 Ω, 1 mm long resistor will have a higher electric field in it than a 10 Ω, 100 mm long resistor, because the resistance per unit length is higher.

I think you're mostly right about needing a higher electric field to push charges through a greater resistance, but you need to make sure you're talking about the right quantity. In this case, it's resistivity or resistance per unit length, not resistance. You're just going to get confused if you're mixing the electric field (a property describing a single point along the circuit) with resistance (a property describing a whole section of the circuit).

 

[sophiecentaur]

So a longer resistor of the same resistance doesn't produce a stronger e field in fact the opposite a weaker one if measured in a single point in the resistor compared to that of the shorter resistor, but since the distance is longer the overall field exerted force on charge is the same, because a given PD has only a given amount of force which cannot get bigger by itself right ?It kinda seemed like that, following my inner logic which is not always the best one ofcourse.

Oh and by the way , if I have a circuit with a wire of negligible resistance and a resistor of 100ohm in the middle then the field in the resistor will be larger as compared to that of the wire which had very low resistance?
It kinda seemed like that, following my inner logic which is not always the best one ofcourse.

I guess upon this recent misunderstanding you missed out on the other quesion, about the battery terminals and extended length of wire attached to it and the volts per metre thing?
last paragraph of my previous post. :)

Resistance doesn't "produce" anything. Would you say that Mass Produces Acceleration when a force acts on it?
It is a shame that you would rather insist on approaching this stuff on your own terms, rather than going through it is a tried and tested way. Unless you are really really clever, you cannot hope to get to the end of it all successfully as you will have to more or less invent your own Science. Very few people have managed to do anything like that - and all the 'greats' had the sense to learn it all the conventional way first. You really don't know what you're getting yourself into, I think.
Do you not think it odd that the 'Potential way through' has been used successfully by so many well informed people?
You have already made a load of pitfalls for yourself and you've managed to puzzle yourself, even at this stage. Why not give in and do a bit of regular 'homework' on the subject? You won't regret it.

 

[Mr.Bomzh]

@Sophie It's not that I'm fighting for a new way of learning science nor am I in favour of some jibberish explanations around fields and stuff, it's just that we each approach the same thing differently , I just ask the questions which then get me the best understanding, I am reading up on some pretty basic PD, volts, and Ohm's law type of things , but I also am interested in stuff from a deeper perspective, ofcourse it brings some confusion along but taht's why I never accept anything before I haven't got precise and trusted answers.:)
And thanks for helping me out.
By the way that resistance produces stuff was a badly worded thing from my part, I never thought resistance produces e field ofcourse not , I should have better said resistance in a given area as I am now explained to say causes the field to get higher due to the fac that in that particular place it is harder for the current to go through because of the higher resistivity.

Ok so by resistance we usually refer to a whole circit or a device etc, but when we want to examine a circuit which has two or more parts that each have different values of resistance we say that this part of circuit aka the resistor has more resistivity than the wires connecting that resistor to the battery terminals for example is that correct?

And last that 10 ohm 100mm long resistor has lower field at any given place along it's length because if we divide the length 100mm by the resistance of the resistor which is 10 ohm we get that every 10mm of the resistor corresponds to 1 ohm of resistance.So here 10mm has te same resistance as 1mm of resistor in your example and that's why the field per given place is lower is that correct?

 

[thegreenlaser]

Resistance doesn't "produce" anything. Would you say that Mass Produces Acceleration when a force acts on it?
It is a shame that you would rather insist on approaching this stuff on your own terms, rather than going through it is a tried and tested way. Unless you are really really clever, you cannot hope to get to the end of it all successfully as you will have to more or less invent your own Science. Very few people have managed to do anything like that - and all the 'greats' had the sense to learn it all the conventional way first. You really don't know what you're getting yourself into, I think.
Do you not think it odd that the 'Potential way through' has been used successfully by so many well informed people?
You have already made a load of pitfalls for yourself and you've managed to puzzle yourself, even at this stage. Why not give in and do a bit of regular 'homework' on the subject? You won't regret it.

I don't think there's anything radically unconventional about what he's asking... the point form of Ohm's law, J = E/ρ is pretty well known. Besides, he's trying to understand a paper which specifically referenced the electric field strength within a resistor.

That said, I agree there's a reason we like to use the bulk quantities (voltage, current, and resistance) rather than the point quantities (electric field, current density, and resistivity/conductivity) whenever we can. When you talk about a resistor in terms of voltage, you're only worried about three numbers: voltage, current, and resistance. When you talk in terms of electric field, you need to know three quantities at every location in 3 dimensional space, which is much more difficult to deal with. However, in this case where he's trying to understand a paper which specifically mentions the electric field, I'm assuming we're pretty much stuck using the electric field approach.

To simplify the analysis, though, we can probably get away with using a 1 dimensional electric field approach rather than the full 3 dimensional approach. Basically assuming that the electric field and current are always parallel to the wire and that the only important thing is how the electric field and resistivity change as we move parallel to the wire. That's basically what I did in my post above, and the 1D approach is really not that hard to understand. The 1D point form of Ohm's law is just I = E/r where r is the resistance per unit length. E and r both change as functions of position along the wire while I is constant. From that, it's easy to see that in places where r is higher, E must be higher as well since I is constant.

The thing is, whether you do a 1D or 3D electric field approach, "resistance" is meaningless if you don't specify how that resistance is distributed. I think that's why the OP is getting confused. He keeps talking about "resistance" in the wrong context. If he wants to talk about the electric field, he needs to, first of all, specify exactly how that resistance is distributed in space, and second of all, either talk about resistance per unit length (1D) or resistivity (3D), because those are the quantities which are directly tied to the electric field.

 

[sophiecentaur]

@thegreenlaser
I wonder what you learned first. I'd bet it was Volts, Amps and Resistance. Same for me and for pretty well everyone else who reads this. What you've written above was coming for some degree of knowledge- enough to see the parallels between the two worlds and you can see where one or the other approach is 'best'. It's one thing to appreciate where the two approaches come together and it's an entirely different thing to try to get circuit theory from the direction of fields.
That's why I am doing my best to encourage MrBomhz to get to his goal of understanding in a way that's been proven to be fruitful. I have to assume that he's not a budding Einstein or Hawkin and that he could do it all from scratch - but then again, neither did those two. They went into it from the 'normal' direction. You have pointed out enough disadvantages in trying it the other way.

 

[Mr.Bomzh]

I think everybody goes in this life and studies the normal direction it's only that people like Einstein and others had a big capacity and also not less important a different view, like a photographer , anybody can learn how to take a picture but not anyone has the mindset to see the right angle that looks the best and when a profesional takes a picture of the same thing you just took it looks very different.
That said I'm not trying to be Einstein or somebody else I'm my myself and sometimes I kinda learn physics upside down, I'm not saying it's better it's just that we are different I just have a kinda abstract mindset, a lot of people have told me that , everytime I tried to get into this normal picture of how things should be I failed even more than without it.Even though I happily agree that the most fundamental things should be learned the best and easiest way possible.On the other hand it's not that Ia hvent read about ohms laws or measured transistors or resistors with an little chinese made analog multimeter and other stuff that I ahve did to build a few circuits I have done.
This time I wanted to get a little deepr for a purpose , I'm sure Sphie and others will see the purpose later.
I'm just going through some things I already know and some that I don't know yet ,

Oh by the way can someone answer the questions I gave in my last post they kinda got behind this " how to teach" debate.
anyway from my opinion your both right so , everything's cool.:)

 

[sophiecentaur]

Oh by the way can someone answer the questions I gave in my last post they kinda got behind this " how to teach" debate.
anyway from my opinion your both right so , everything's cool.:)

The easiest way to work that out is to work out the voltages at the junctions between the resistor chain; the usual way*, which you can rely on to work. Once you have that, you could work out the fields. Personally, I just wouldn't bother doing it the other way round - far to much trouble.

*http://www.bbc.co.uk/schools/gcsebitesize/design/electronics/calculationsrev2.shtmlt.

 

[thegreenlaser]

I missed your last post, sorry. I will say, though, if you're trying to learn circuit theory via electric fields, then I agree with sophiecentaur... that's the wrong way to do it. However, since your question was specifically about the electric field strength inside a resistor, I assumed that you're already comfortable with both circuit theory and electric fields, and you just want to connect the two. A full field analysis of circuits is suprisingly complicated (you would probably need to use a computer simulation), but your particular question isn't so bad if you have a basic understanding.

Ok so by resistance we usually refer to a whole circit or a device etc, but when we want to examine a circuit which has two or more parts that each have different values of resistance we say that this part of circuit aka the resistor has more resistivity than the wires connecting that resistor to the battery terminals for example is that correct?

Typically, yes, but high resistance does not necessarily mean high resistivity. To find resistance from resistivity, you also have to know the shape of the object. A very long, very thin copper wire might have a higher resistance than a short block of wood, even though the resistivity of copper is much lower than that of wood.

And last that 10 ohm 100mm long resistor has lower field at any given place along it's length because if we divide the length 100mm by the resistance of the resistor which is 10 ohm we get that every 10mm of the resistor corresponds to 1 ohm of resistance.So here 10mm has te same resistance as 1mm of resistor in your example and that's why the field per given place is lower is that correct?

Resistor A = 1 Ω, 1 mm
Resistor B = 10 Ω, 100 mm

A corresponds to a resistance per unit length equal to 1000 Ω/m, while B corresponds to 100 Ω/m. So A has a higher resistance per unit length, and thus will have a higher electric field, no matter what the current is, because (electric field) = (current)*(resistance per unit length)

Here's yet another way to look at it...

Say there's a 1 A current flowing through both resistors. Resistor A will have a voltage drop of 1 V across it. A 1 V drop over 1 mm gives an electric field of 1000 V/m since (electric field) = (voltage drop)/(distance). Resistor B will have a voltage drop of 10 V by Ohm's law. A 10 V drop over 100 mm gives an electric field of 100 V/m, which is smaller than the electric field in resistor A.

So there's two ways. You can directly find the electric field using the resistance per unit length, or you can find the voltage using the resistance, and then find the electric field using the voltage. Either way, though, you need to know the lengths of the resistors to find the electric field, which is why people usually only bother to find the voltage unless they really need the electric field.

 

[sophiecentaur]

@MrBomzh
I reckon the best thin you can do is work through some basic exercises with simple resistive networks and take it from their. From what you say, I gather that you haven't done a lot of that stuff. I think if you had, you'd not be so fixated on the other approach. Your way is certainly not the easiest way into it. For a start, you won't find any support, whereas, the conventional way will make a vast amount of support material at any level to suit you.
That is if you really want to get to grips with the subject.

 

[Mr.Bomzh]

Ok so if I get it correctly , if we increase the resistance keeping the length te same the field increases, like the resistor B had only 100v/metre because it was 100mm but if it would be 10mm it would be E=10/10 =1, just as A , or maybe E=10/1, what's then ? keep the 10 ohms of resistance but the length is now 1mm , now I get an answer doing 10/1 which is 10, but what does that 10 stand for, one added zero to the field strength ?

Ok I have to lay down my cards , so that you folks don't get confused for my questions.Ofcourse I agree that learning electric fields for trying to understand circuits is not a handy thing to do and I don;'t need this for that.Ofcourse nobody builds a cirucit and calculates the electric field of a resistor , it doesn't matter as long as you know the voltage/current drop it's fine.
It's just that I have a different plan in mind I'm not asking this for a circuit or so much for the basic learning stuff, I'm asking this more for an idea or ideas I ahve in my mind which are kinda complicated so I can't solve them always myself.
I'm sure Sophie kinda knows what I'm talking about since he participated in my previous threads.
the idea is more about that I need a high electric field from low voltage in a small area , which would be surrounded by a dielectric material , like a current path in the middle of a dielectric , but ofcourse it has to have high resistance so that I wouldn't need much current, yet i need the e field.
I was just wondering what would happen if a dielectric placed between the plates of a capacitor had a resistive layer inbetween which would be attached to a circit and current would pass in it , we know that the charges in the dielectric align themselves according to the charges on the plates oppositely ofcourse.
I was just wondering what would happen to the alignment of the dielectric charges if the dielectric had a e field inside of it.
Just an idea so please be nice on me :D

 

[sophiecentaur]

Ok so if I get it correctly , if we increase the resistance keeping the length te same the field increases,

One thing at a time. If you have the same PD across it, changing the resistance makes NO difference at all to the field. Unless you get the right variables in your mental equation, you will fall over. Use the formulae and don't put the cart in front of the horse. The amount that the PD increases will depend entirely on the other resistances in the circuit - which is why I say you should work it all out the proper way. You are wandering around in a miasma of mis-information. Get a proper grip on one thing at a time and you will start to make headway.

 

[Mr.Bomzh]

Well I was speaking about a case where there are no other resistances, or the ones that are are negligible for the sake of simplicity , like a resistor connected to a battery.
Now given the example , what happens with the field , Ok I understand that the overall field stays the same as it is related to the PD and cannot just increase as that would also imply an increase in energy which would relate back to a need for an increase in PD, I'm speaking about the field for a give place ?
So in the situation that a described is it so that given fixed PD and fixed field strength decresing the length of the resistor but keeping everything else constant will increase the field in that particular place in the resistor?

Atleast it seems so when I plug in the varyables into the formula you guys gave me.
E=length/total resistance of resistor, now E=10/10 which is 1 yet if I amke 10/100 it's 0.1, well I think i got it , if this is the right way to calculate it ofcourse.

 

[sophiecentaur]

Every one of your posts seems to contain confusion. Your "plain English" is not even plain English (scientifically speaking).
I can't imagine how you can think the Field is length / resistance when it's PD/ distance. But from your self-imposed position, it is not surprising that you are not getting anywhere. Just give in and do it properly.
I give up.
Come back when you are serious about learning some serious Physics in the right way.

 

[Mr.Bomzh]

No i didn't mean that field is length /resisatnce ofcourse it;s PD/distance , just as the law " volts per metre" points out.
But just like in the water analogy , even though the smaller pipe (resistor) is located further from the source , it still has higher pressure in it than other parts of the pipe due to the fact that it's cross section is smaller in diameter so water builds up pressure to get through it so that the overall flow would remain the same.

My question was about a given fixed PD say 100volts and a resistor in that circuit , nothing more just a resistor , assuming the leading wires to the resistor have negligible resisance and the field in them is lower but since the resistor has a higher resistance the field in it builds up so that it could keep the current flowing through it due to the fac that it takes more power to move the same amount of currrent through that particular part of cirucit than the other ones , which in my case were low resistance wires. Is this really wrong , tha thing i just said?
even though it seems logic just as for the smaller diameter pipe to haev more pressure to get the water through as for thre resistor to have stronger field than the wire right before it?

 

[sophiecentaur]

Seriously, I really don't want this sort of conversation. What you are writing is not in the spirit of PF. You are just trying to force the Physics of the situation to suit your particular naive view. Consequently, you are making hardly any progress. Until you are prepared to talk in the accepted terms - using Maths, where appropriate, to show you understand what you are saying - I don't want to carry on this conversation. In 23 posts, we have got nowhere at all.

 

[Mr.Bomzh]

Another quesion I wanted to ask and I already did but didn't received an answer was this, the electric field said as volts per metre , from which point does it start to fall of as 1/r2 according to the law, does it satrt to fall of from the battery + terminal to which the wire leading up to the resistor is connected or dopes it start to fall off from the resistor or where?
It would be easy to understand this for a point charge , since the field starts to fall of directly from the charge , but when a wire that is connected to the + terminal of a battery or any power source for example is connected can we say that wire then becomes just an extension of the positive terminal so the field should be just as strong at the potential as it is at the end of that wire attached to the terminal?
So if this wire is attached to a resistor does that mean that the field starts to fall off only at the resistor ?
I attached an image to illustrate what I mean.
I also made numbers at different part of the circuit to show and ask what is the field strength (approximate) there. So my guess is that at number 1. the field has it's maximum value given for the PD the batery has , say 12volts.
At number 2. this is the one I am not sure , its either the same as at number1. or less , depends on the answer to the quesion I gave in thsi post.
at number 3. it's dependant on the relationship between PD across the resistor, it;s length and total resisatnce divided by that length to arrive at a field strength at a particular point on that resistor is that correct?
at number 4. it's low , dropped due to the decrease in potential over distance or in other words it's lamoust close to ground, negative terminal.

Ok I undersatnd Sophie, well you are welcomed to give me an answer in terms of PD , volts etc use a formula if you need to I will try to understand it if it's not too complicated for me

I guess the problem is that I;m not good at maths, I think I understand the idea It's just that my wording and level of expression is unclear or not suited to some particular stuff, maybe if I could express this in some formulas and equations you or others could understand better

 

[Mr.Bomzh]

Well according to what @Zapperz said yesterday in his post , there should be no overall difference in the electric field in a resistor of given ohms say 100 and given PD , the only reason that a shorter resistor could have a higher field strength in a given point along it's way than a longer one is because the same field due to the same resistance and same PD now has to occupy a smaller/shorter space , this should be correct? I agree to what Zapper said and aslo to what you sophie said it;'s just that I wanted to know the reason behind this. Comparing this to the inclined lines of Zappers post and the car running uphil to what you Sophie said I can conclude that even though the energy used by the car to get to the same potential hegith is the same in both cases , in a longer climb the energy used per a unit length is smaller due to the unit length being longer itself right? the opverall energy is the same but the energy per unit or the field per given point is stronger, my inner voice tells me I should be correct about this one. I hope so.

 

[sophiecentaur]

Perhaps you should tell your "inner voice" to shut up and to give your brain a chance at this problem.

 

[Mr.Bomzh]

I do need to correct that piece of idiocy in my last post. Of course, increasing the length of the resistor Decreases the field (the Volts per Metre).

Actually I start to find out answers myself now, looking back at the posts you Sophie said this so if increasing the length decreases the field per metre then the opposite must be true that decreasing the length but keeping the same resistance of the resistor should increase the field , volts per metre. I still hope to get some light on my previous question posted in the last two posts excluding this one.

 

[thegreenlaser]

Well I was speaking about a case where there are no other resistances, or the ones that are are negligible for the sake of simplicity , like a resistor connected to a battery.

Woah, hold on. This is a completely different problem. The underlying assumption through most of this thread has been that you're comparing resistors which have the same current running through them. I.e., you're comparing the electric field in two resistors which are in series. If you have two resistors in series, then the electric field will be strongest in whichever resistor has the highest resistance per unit length. However, that's only because both resistors have the same current running through them. As I said before, the relevant formula is:

electric field = current * (resistance per unit length)

If you hook one resistor up to a battery, calculate the electric field, and then hook up a different resistor to that battery, that's a different question altogether, because the resistors won't have the same current flowing through them. You would have to find the current by Ohm's law, and then calculate the different electric fields using the formula above.

Given your posts so far, inclined to agree with sophie that this resistance per unit length approach may not be best for you. If you want to find the electric field strength in a circuit, here's what you should do.

1) Solve the circuit. Use Ohm's law to find all the voltages across all the resistors.
2) Find the electric field in a given resistor using: (electric field) = voltage/(length of resistor)

Then you can compare the electric fields as needed.

It's just that I have a different plan in mind I'm not asking this for a circuit or so much for the basic learning stuff, I'm asking this more for an idea or ideas I ahve in my mind which are kinda complicated so I can't solve them always myself.
I'm sure Sophie kinda knows what I'm talking about since he participated in my previous threads.
the idea is more about that I need a high electric field from low voltage in a small area , which would be surrounded by a dielectric material , like a current path in the middle of a dielectric , but ofcourse it has to have high resistance so that I wouldn't need much current, yet i need the e field.
I was just wondering what would happen if a dielectric placed between the plates of a capacitor had a resistive layer inbetween which would be attached to a circit and current would pass in it , we know that the charges in the dielectric align themselves according to the charges on the plates oppositely ofcourse.
I was just wondering what would happen to the alignment of the dielectric charges if the dielectric had a e field inside of it.

Any standard course in electromagnetism (particlarly one with an engineering focus) should answer all these questions. Parallel plate capacitors with different layers of dielectric/conductive materials are pretty commonly seen in such courses.

Another quesion I wanted to ask and I already did but didn't received an answer was this, the electric field said as volts per metre , from which point does it start to fall of as 1/r2 according to the law, does it satrt to fall of from the battery + terminal to which the wire leading up to the resistor is connected or dopes it start to fall off from the resistor or where?

The 1/r² law applies only to point charges. A circuit is not a point charge.

You might think that the electric field should get weaker farther away from the battery, but it turns out that there's actually a feedback mechanism which sustains the electric field away from the battery, making it not look like a dipole anymore. I won't say more than that because it gets really ugly really fast (there was a thread on this a little while ago). Thinking of the battery as a dipole really isn't prudent in this case. You're best off just trusting circuit theory here. Find the electric field by calculating voltages, not by thinking of the battery as a collection of point charges.

Try finding the electric fields in your diagram again, but this time do what I said above. Find the voltages across each of the elements: first the wire, then the resistor, then the other wire. Divide those voltages by the length of each element (wire, resistor, then wire) to get the electric field inside those elements. (Assign some small resistance to each of the wires to make things easier). What do you find?

Hint: the electric field doesn't get "used up" as you go along the wire.

Well according to what @Zapperz said yesterday in his post , there should be no overall difference in the electric field in a resistor of given ohms say 100 and given PD , the only reason that a shorter resistor could have a higher field strength in a given point along it's way than a longer one is because the same field due to the same resistance and same PD now has to occupy a smaller/shorter space , this should be correct? I agree to what Zapper said and aslo to what you sophie said it;'s just that I wanted to know the reason behind this. Comparing this to the inclined lines of Zappers post and the car running uphil to what you Sophie said I can conclude that even though the energy used by the car to get to the same potential hegith is the same in both cases , in a longer climb the energy used per a unit length is smaller due to the unit length being longer itself right? the opverall energy is the same but the energy per unit or the field per given point is stronger, my inner voice tells me I should be correct about this one. I hope so.

If the voltage drop is the same, then the electric field is just the voltage drop divided by the distance over which that voltage is dropped (the length of the resistor in this case). So yes, shorter distance would mean stronger electric field.

Again, though, make sure you understand which values are being held constant and which ones aren't. If the current is constant, that's one thing, but if the voltage is constant that's another thing. If the voltage is constant, then the resistance has no effect on the electric field. If the voltage is constant, the only thing affecting the electric field is the length over which that voltage is dropped.

I really think you should try to take a good course on electromagnetism if you want to understand this stuff better. What started out as a pretty simple question seems to be turning into you needing explanations of a lot of things which would all be covered in basic EM/circuits classes. I don't mean to reject your questions, but the explanations you need are far more than a single forum thread can offer.

 

[sophiecentaur]

If you hook one resistor up to a battery, calculate the electric field, and then hook up a different resistor to that battery, that's a different question altogether, because the resistors won't have the same current flowing through them. You would have to find the current by Ohm's law, and then calculate the different electric fields using the formula above.

.

The current doesn't affect the 'Field'. It is just the Volts per Metre that counts. If a battery is supplying the volts then they are what they are, with or without a resistive path.

Bad idea to use the term "Feedback mechanism" in this context. Feedback, in Science implies the existence a control loop with active amplification in the feedback path.

I totally agree with your idea that he should take a proper course about this. Privately inventing a private version of Science for yourself seldom proves a fruitful exercise. No pain no gain MrB.

 

[Mr.Bomzh]

Well my try on the field strength calculation would go like this assuming both wires leading up to the resistor from the battery terminal have something like 2 ohms os resistance and they are say 10cm each, assuming 12v on a typical battery that would be like E=12/10 which is 1.2, although it's kinda strange because I just divided the resistor length by voltage but shouldn't this also include the total resisatnce of that particular resistor?

Or maybe I have to do the above formula first and then put the number into the second formula which you actually posted first which goes like E= Ix (resisatnce per unit length) from the previous formula I just found out that the number for resisatnce for unit length is 1.2 so now I will attempt to calculate current which is I =V/R, now I= 12/2 which is 6amps of current.

So E = 6 amps x 1.2 which is 7.2, volts per metre I guess?

Now as for the resistor , assuming it is 100 ohm and 4cm long , E=12/4 =3(resisatnce oper unit length)
then I=12/100= 0.12mA , now E=0.12x3 =0.36 again volts per metre?

So in this case the field in the resistor is smaller than that in the wire?
Also in a single resistor case , I guess the total current that flows from the battery + to - is not the current I calculated through the 2 ohm wires but the current of the 100ohm resistor because in a series circuit current through each component is the same and the overall current in a series circuit is the current of the part with the highest resistance which in my case is the 100ohm resistor is that correct?

 

[thegreenlaser]

The current doesn't affect the 'Field'. It is just the Volts per Metre that counts. If a battery is supplying the volts then they are what they are, with or without a resistive path.

Current is related to electric field... E = V/L = I*R/L and J = σE are both relationships between current and electric field which are valid in the context of this thread. However, I agree that E = V/L is the more useful way to go in the case of constant voltage. I've had to backpedal a bit because I gave an answer specifically targeted at the original question (series resistances) and then that answer started getting eroneously applied to a different context.

As I said before, since we're now dealing with electric fields in general circuits, I agree that E = V/L is a better way to go.

Bad idea to use the term "Feedback mechanism" in this context. Feedback, in Science implies the existence a control loop with active amplification in the feedback path.

I'm well aware of what a feedback loop is, but maybe I'm wrong about how the electric field gets set up in a circuit. My main intention, though, was to convince MrB that it's really not worth opening up that can of worms, especially without a lot of experience with EM theory.

I totally agree with your idea that he should take a proper course about this. Privately inventing a private version of Science for yourself seldom proves a fruitful exercise. No pain no gain MrB.

Glad we agree.

@MrB: You have 3 resistors in series connected to a battery, so having a 12 V drop across each of them would be a direct violation of Kirchoff's voltage law. To be blunt, this is a really simple circuit and the fact that you're assigning 12 V to each element tells me that you don't really understand basic circuit theory.

I don't mean to belittle you; your enthusiasm for this stuff is great, but you're getting in over your head. You need to start with the basics of circuit theory and electromagnetism before trying to understand how they align with each other. You'll be much more satisfied with your understanding that way.

 

[Mr.Bomzh]

with my calculations I concluded that this current has to be 0.12mA, is that correct?
Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

I did made a mistake previously I should have just counted the wire resistanes in with the resistor because they altogether make one large resistor just with different resistivity along the way , smaller along the wire and higher along the resistor.

But my use of formulas was correct ?

So in a a series circuit with two resistors, one 4 ohm other 8ohm and a 12v battery, the current through each would be the same but voltage drops different ,For the current in the circuit I= V/R so 12/12 = 1 amp. of current , for the 4 ohm resistor V=1x4 and for the other resisor V=1x8, so the question is the current is the same through both resistor but the resistance of each differs so the electric field in the 8 ohm resisor is higher than the field in the 4 ohm resistor , assuming both have equal lengths? Ad the reason why it is higher in the 8 ohm resistor is because to force the same current through a higher resistance needs more emf which relates to higher voltage?
I also use this page to read about resistors , find it good
http://physics.bu.edu/py106/notes/Circuits.html

 

[sophiecentaur]

Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

But you started with a 12V battery, so it is hardly surprising that the voltage across it is 12V. If it were not, you would need to take it back to the shop!

 

[Mr.Bomzh]

Yes , I get it for a single resistor over the battery the voltage drop is always going to be 12 volts , because upon measuring with a voltmeter , you basically attach to the + and - leads of the battery which also happen to be the wires entering and exiting the resistor. a two resistor case would be more interesting , i made some calculation in the previous post why doesn't anybody say something about them ?

thank you . :)

 

[sophiecentaur]

Forget the meter. A battery is a Voltage Source - period.

 

[thegreenlaser]

with my calculations I concluded that this current has to be 0.12mA, is that correct?
Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.

I did made a mistake previously I should have just counted the wire resistanes in with the resistor because they altogether make one large resistor just with different resistivity along the way , smaller along the wire and higher along the resistor.

But my use of formulas was correct ?

I don't think your use of formulas was completely correct, but honestly I got pretty lost in what you were doing so I don't know what exactly you did wrong.

So in a a series circuit with two resistors, one 4 ohm other 8ohm and a 12v battery, the current through each would be the same but voltage drops different ,For the current in the circuit I= V/R so 12/12 = 1 amp. of current , for the 4 ohm resistor V=1x4 and for the other resisor V=1x8, so the question is the current is the same through both resistor but the resistance of each differs so the electric field in the 8 ohm resisor is higher than the field in the 4 ohm resistor , assuming both have equal lengths? Ad the reason why it is higher in the 8 ohm resistor is because to force the same current through a higher resistance needs more emf which relates to higher voltage?

As far as I can tell, this paragraph is correct. You should have done something more like this with the 100 Ω resistor and the 2 Ω wires. Find the voltage across each resistor exactly like you did here. Then, once you have the voltages, divide those voltages by the lengths of the elements to get the electric field in volts per metre.

 

[sophiecentaur]

I already referred him to the potential divider theory but it must have been too straightforward for him to bother with.

 

[thegreenlaser]

Indeed...

Well, I hope what's been written in this thread will become more useful once MrB takes some time to learn basic circuit analysis and EM. That's what the rest of us had to do too...

 

[Simon Bridge]

To me, this seems to be the heart of the problem:

Ok I have to lay down my cards , so that you folks don't get confused for my questions.

That is a very good idea.
However, I don't think you are being entirely straight with us even so... see below:

the idea is more about that I need a high electric field from low voltage in a small area , which would be surrounded by a dielectric material ... it has to have high resistance so that I wouldn't need much current, yet I need the E field.

That's situation 1.

I was just wondering what would happen if a dielectric placed between the plates of a capacitor had a resistive layer in between which would be attached to a circuit and current would pass in it , we know that the charges in the dielectric align themselves according to the charges on the plates oppositely of course.

This is situation 2.
It is very different from the situation in the last passage I quoted.

I was just wondering what would happen to the alignment of the dielectric charges if the dielectric had a e field inside of it.

This is an different situation again ... so you are, in fact, pondering three distinct situations.

Interestingly, circuit theory is not helpful for any of these situations.
Much confusion stems from trying to overlap two different ways of understanding something.
Stop it.

In situation 1: you need a short length of high resistivity material.
There are material constraints about how short the length can be and so on that you will discover by experimentation.

In situation 2: you are describing something like a carbon rod passing perpendicularly between two plates. "wonder what would happen" is too vague though - lots of things happen - you need to ask a specific question. What sort of effect are you interested in?

eg. electric fields (i.e. due to the capacitor) that are perpendicular to a wire (or resistor etc), have no effect on the electric current or voltages along the wire.

Situation 3: this is different from situation 2 because now you want an extra field in the dielectric (in situation 2, there is a rod of carbon where the extra field is - the dielectric is outside).
This would result simply from using two sets of plates.
The polarization of the dielectric is simply aligned with the combined field.


You still have not been straight with us though: what is the application you have in mind?
What are you hoping to discover?

 

[Mr.Bomzh]

To Sophie, excuse me I did not ignored the things you gave on purpose , it's just that sometimes I miss out on some stuff you either say or give as I am too concerned about the context, I will look back on the posts as I do to reread them and understand better some stuff.

To Simon and to others , as I said my writing is confusing and vague amny times not so much because I don't undersand but because while trying to explain myself as best as I can I'm gettin nervous and stresses and my thoughts don't go together with how I express them, although there are also pure misunderstanding by my part.

And finally , well what am I trying to doscover? I have a feeling and some thoughts what it would be but I don't have the full idea of will it work. Well you see I can repair a old tv if it's not extremely complicated and I ahve done a few, I can repair a amplifier and I understand some basic stuff, it's just that I'm a lonely person sitting at my desk and having great love towards all these devices and how their work , that love and passion is affected though by my lacking maths and my mind which tells me that I'm just an ordinary little repairguy who will probably undersand just as much as all the other grey people from the huge grey masses of society.A very suicidal feeling that is I have to tell you. :D Well lacking maths but I do have one thing , I ahve the ability to visualize thing very deeply, So i;m sitting reading papers painting pictures in my head of how those phenomenon work , when I was smaller I figured out the internal combustion engine not so much from books but froms people telling me how it works and me visualizing how it works , and a few years after that I built one , in other words I took an old one from my grandfathers old car and completely restored it all new and working and put back in.There were people helping me with advice ofcourse, some old mechanics that I made very good friends with afterwards.
So with electronics it's the same , I sit read papers try to figure out stuff, I look at what's built and so on and so I came to this capacitor idea, I was thinking like for half a year non stop how could one arrange a cirucit which would use capacitors for purposes like the ones I describe.
Maybe I hate transistors , mosfets in smps applications especially , because I have never had good luck with them because it is not easy to build a device with them you have to know everything your doing very good, it's not like building an old combustion engine , I would say it's much harder in terms of intellectual power needed,
I mean I tried both and I failed at the latter one.Even though it was a really simple smps but then again a simple smps is a easy to destroy smps at the same time, taht's why I had this idea about instead of driving a transformer with direct current paths being reversed in polarity each cycle making a varying flux in the core, using electric fields to attract , repel charges which would form varying current through a primary cirucit.
i do believe , from what I know , that it;s possible , just not entirely sure of how.

Excuse me for my long memoirs , but you wanted to know what am I after and I told you.
Will write less and more precisely later one when I'm home about the progress I ahve made with you guys and the idea itself.

 

[Simon Bridge]

smps is a complicated power supply cicuit - to learn transistor circuits you have to start a LOT simpler than that.
Usually just start with a switching circuit and/or an amplifier. Maybe a flip-flop. Basic stuff to get a feel for what they do. There's a lot of resources with things for the hobbyist to try.
i.e. http://www.pcbheaven.com/wikipages/Transistor_Circuits/

The thing about mechanical engines is that the parts are big enough to see what's going on.
Electronics is a lot more spooky.

Learning about transistors is probably harder now than when I started - so much of what I used to do with them has been taken over by integrated circuits. You may like to look at simple FM receivers though.

The variable capacitor thing requires some sort of external work. You cannot get away from that.
Until you have the basics down, I don't think we can tell you any better than that.
Perhaps you'll just have to try and see.Here is the main problem for us: your goals, as you are describing them, are far too vague for anybody to really help you with them. The kind of things you want answers to keep shifting about, so nobody knows what to tell you that will do any good. We just keep guessing in the dark - that's no good to you.

I think you need to have a sit and think what you want to achieve - specifically.
The take away lesson for you in this is to keep circuit theory separate from electric field theory.
Don't mix them up.

 

[Mr.Bomzh]

Well basically , as I said earlier I wanted to make a both polarity and amplitude varying voltage source that could be fed into the transformer pirmary in order to then convert it to whatever voltage /current needed + the isolation part between primary and secondary that the transformer gives.
The transformer is the long known part and it works just okay.The thing is to make it work with dc power source.We have smps and all kinds of other topologies etc.
I was just thinking of a easier way to do it, not saying it possible just thinking, learning in the process etc.

So we concluded after all these two thread talk that a capacitor in a circuit being in series with the pirmary could pump charges back and forth through the pirmary if it's capacity could be altered , the problem here ofcourse is how does one do that , not speaking about the rotors or any mechanical stuff but in a solid satte way.

A mechanical analogy here , when I was younger i remeber an abandoned building which had elevator shafts on the top floor in the elevator engine room , there were these concrete pads on which the electric motors that pulled the cables were located, these pads stand on 4 springs , I stood on the pads and jumped and after a few jumps I could make them oscillate, now if I would just stand there doing nothing the pad would also stand still, DC is like that it just exerts a constant pressure so to speak but nothing happens , now imagine these pads and the weight on them , if one could alter the stifness of the springs while I stand there , the pads would start to oscillate just as if I would be jumping on them , only in this case I wouldn't be jumping rather standing still but altering the stifness of the springs would make the pad go lower and then higher again and so it would gain inertai and try to reach it's resonant frequency probably , which would also be largely dependant on the rate at which one altered the springs.

In my capacitor transformer case it is similar , the dc source applies a constant voltage to the circuit but altering the capacitors capacitance with the help of the dielectric one could make this oscillation in the cirucit.
Im not sure not an expert on materials science but maybe there is a nanotechnology way to make a dielectric whose dielectric constant could be altered in a large enough amplitude to cause noticable changes in the capacitance of a capacitor, ofcourse that also begs the question by what method the dielectric could be made to change , an applied voltage or a field or whatnot.

Speaking about dielectrics I have been looking and what happens to their inner workings when they are in a capacitor and the PD is applied across the plates to charge it up.Diagrams show that polarization happens inside the dielectric as molecules polarize themselves to allign according to the applied field from the plates , what would ahppen if one could alter this alignment process inside the dielectric material ? how would that affect the field between the plates, now we know that the dielectric itself affects the field between the plates , it decreases it's strength which is the reason more charge has to flow to the plates to set equilibrum condition.

 

[sophiecentaur]

Hi again. I thought you'd given up!

I'm afraid that I can't drag myself through all that stuff you just wrote because you are still missing the main point of all this. You need to supply all the energy for this scheme of yours in the system you will use to change the permittivity because none will come from the battery in your circuit diagram. There is nothing more to discuss until you accept that. If you can't realize how fundamentally flawed your idea is then you will get nowhere.
What is the point of trying to invent a hypothetical material to do a job that can't be done, your way, in any case?

 

[Mr.Bomzh]

Yes I haven't and I never will:D
I'm not attempting a PMM machine I think I made that clear in the beginning, yes ofcourse there will have to be some energy transfer involved to do the job.

Ok why not rather speak about the mechanical analogy, far easier there yet the basic principles apply.
I stand on a cement pad which is suspended in air by 4 springs.The springs are compressed because the weight is pushing on them constantly , so this is a steady situation , nothing changes , given amount of weight given amount of spring stifness.
Now there are two ways to achieve motion of the concrete slab , one way is to jump on it which would create oscillations up and down , by jumping I would apply force each time I hit the slab when I land on it.So the energy in this scenario comes from my jumping/muslces etc.
Another way would be as I said altering the stifness of the springs.In this situation ok I'll be honest it is kinda tricky to find out where the energy comes from.but it definitely comes from whatever jamms the springs or from the springs themselves depending on what kind of process or material comes into play here.
Iamgine springs that get like two times stiffer once you reach a certain compression level.Ofcourse if the slab comes down it presses on the springs which then store this gravitational potential energy the slab had before , now this energy stored in the springs could be released and with a little additional energy input could lift the slab to its previous height, is tha correct?

 

[sophiecentaur]

If you don't include a power source then you are proposing a PMM.
I understand exactly what you are saying about the equivalence of spring stiffness and permittivity but both of those things involve input of energy, when the spring / capacitor is stressed. If you do anything to a spring to increase its spring constant (to make it lift something) you will need just as much energy (more,probably, because of efficiency issues) as just lifting the object directly. Your "little additional energy" is a nonsense - it's the TOTAL energy that's needed. You are just trying to sneak in a PMM without realising it. You must realize there is no way round this one. If you stress a capacitor or spring and then use the stored energy to do a job, the next time round you need to supply the same amount of energy again.

 

[Mr.Bomzh]

nowhere did I said there shouldn't be a power source ofcourse there should.the only question is about how it would be connected and implemeted.
ok you say that once the slab or whatever falls and compresses the springs to lift it up would require as much energy input as the first time , but not necesserily so , because while the slab fell and pressed the springs it gave its energy to the springs , assuming some leakage ,and imperfections the next time you would only need to supply the missing force due to those imperfections as the stored energy would release itself lifting the slab, ofcourse I am talking about a situation where the slab is freefalling.If one would attach something to it to do work it would gain less energy and require more energy put into the springs to lift it each next time.
I guess this is the resonance principle , if a girl is thrown from a height H to a trampoline she hits the trampoline and is being thrown upwards by it , only a little less high than the heigh from which she started now putting in some additional energy she could reach the same exact height.I think I'm correct here.

I know that given a primary and a capacitor in series the batery cannot do work , well it can as long as the capacitor is charging and then it stops , a battery can do work to a circuit as long as it can make charge flow, since a capacitor blocks charge flow a battery can do no more work on the circuit.
You see Sophie the idea here isn't about so much whose doing the work , whether the battery when charging the cap or the dielectric when changing the capacitance to make the charges flow back.The idea here is about easier running conditions on the switches.Maybe I should have explained that from the very beginning to make my point better.
In high voltage dc applications a semiconductor switch in series with a transformer like in a half bridge or other topologies has the voltage drop across the switch and due to flux lag or other factors can have to work under very harsh conditions under which it can fail.
By having this special capacitor , I have the idea to achieve the same charge flow as in a directly point to point connect circuit , but only with the help of fields rather than direct current path. Ofcourse the charges still flow through the switch but atleast there isn't a PD across it, instead the PD is across the cap and I think the cap is pretty fine with that , even if the voltage is high.
i hope you get what I am talking about here. :)

 

[sophiecentaur]

So the energy conservation laws apply - but not totally? This is ridiculous. Whatever energy is stored in a spring or capacitor and then used to do work, has to be replaced - all of it. However you re-arrange the argument. Resonance has nothing to do with supplying energy.
You are still not acknowledging the basic flaw in your idea.
The last two paragraphs are just gobbledegook. Science does not work like that.

 

[Simon Bridge]

So we concluded after all these two thread talk that a capacitor in a circuit being in series with the pirmary could pump charges back and forth through the pirmary if it's capacity could be altered , the problem here ofcourse is how does one do that , not speaking about the rotors or any mechanical stuff but in a solid satte way.

No. That is not the problem!
It is a problem but not the main one from the POV of design.
The main one is finding anything that has not already achieved the aim without having to use the capacitor.

Sophie is correct: until you are prepared to acknowledge this basic flaw in the concept itself, you will make no further headway.

 

[Mr.Bomzh]

Oh again my wording problems.Ok I do believe that the energy stored in a capacitor then used to do work has to be refilled back from a power source or whatever kind of external source which is not part of the system of the capacitor and transformer.That's not the problem , never was.It's not about supplying energy , rather making that supplied DC energy ,if one can say like that, oscilate.
The question was about how does one alter the field between the plates.
Thsi is an idea and even if it's impossible theoretically or in practice or both that's not the most important part as I still learn a great deal about the circuit situation and the capacitor since i started this.
What exactly is as you said goobledigook about a capacitor that's charging or discharging used to do work on a inductive device like transformer?

Ok so basically so far I have understood that once the cap is charged up the energy that went into it is now captured there by the charges on the plates and in the electric field that has formed between the plates.
If someone would now want to alter that situation he would have to use some or the same amount of energy to change the dielectric constant and that energy would manifes itself as the change in capacitance and the result would be charge flow?
Ok let's take the mechanical capacitor for an analogy to understand better, say I have a mechanical varyable capacitor , like the old radio tuners.
Say this cap is 100uF and 100v rated.Now the cap is in a circuit situation and is charged up to its 100v and 100uF of charge.That means that the cap now has "x" amount of energy in it stored.If I wanted to drop it down to 10uF by turning it so the plate contact area decreases by 90%, how much work would it require to do so? Would it require 90% of the original amount of energy" x" that was stored in it?
In other words whenever changing the capacitance the amount of work needed to do that is equal to the amount of work you get out from the flowing charge due to the change in the capacitance?