# Electric field, resistor

1. Jan 5, 2014

### 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.

2. Jan 6, 2014

### Simon Bridge

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

3. Jan 6, 2014

### 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'.

4. Jan 6, 2014

### 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?

5. Jan 6, 2014

### ZapperZ

Staff Emeritus
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.

6. Jan 6, 2014

### 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.

7. Jan 6, 2014

### ZapperZ

Staff Emeritus
Thanks.

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

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

Zz.

Last edited by a moderator: May 6, 2017
8. Jan 6, 2014

### 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.

9. Jan 6, 2014

### 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 lenght of wire attached to it and the volts per metre thing?
last paragraph of my previous post. :)

10. Jan 6, 2014

### 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).

11. Jan 6, 2014

### sophiecentaur

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.

12. Jan 6, 2014

### 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 thats why the field per given place is lower is that correct?

13. Jan 6, 2014

### thegreenlaser

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.

14. Jan 6, 2014

### 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.

15. Jan 6, 2014

### 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, alot 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 learnt 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.:)

16. Jan 6, 2014

### sophiecentaur

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.

*This link is a possible start.

17. Jan 6, 2014

### 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.

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.

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.

18. Jan 6, 2014

### 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.

19. Jan 6, 2014

### 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 dont 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 doesnt 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

Last edited: Jan 6, 2014
20. Jan 6, 2014

### sophiecentaur

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.