Doubts about Electric field due to an infinitely long wire

AI Thread Summary
The discussion centers on the electric field generated by an infinitely long straight uniformly charged wire, emphasizing the importance of the wire's infinite length for deriving the electric field expression E = λ/2πε0r. Participants clarify that the electric field is perpendicular to the cylindrical Gaussian surface due to symmetry, which is lost if the wire has finite length, as points on the surface would be closer to one end than the other. The conversation highlights that at finite lengths, the electric field varies along the wire, while for an infinite wire, the field depends solely on the distance from the wire, leading to uniformity in the field's direction. The symmetry of the infinite wire ensures that the contributions to the electric field from opposite sides cancel out any parallel components, resulting in a net field that is perpendicular to the wire. Understanding this symmetry is crucial for accurately applying the electric field formula in practical scenarios.
gracy
Messages
2,486
Reaction score
83
About Electric field due to an infinitely long straight uniformly charged wire my book says the assumption that the wire is infinitely long is very important because without this we can not take vector E to be perpendicular to the curved part of the cylindrical gaussian surface.I think it should be parallel in place of perpendicular or it should be flat part rather than curved part.
I also want to ask there is one more sentence that I am not getting
the expression of E derived from an infinitely long straight uniformly charged wire i.e Vector E= λ/2πε0r
is true largely at the central portion of the wire length and not really true at the ends of wire.I want to know why this is so?
 
Physics news on Phys.org
gracy said:
About Electric field due to an infinitely long straight uniformly charged wire my book says the assumption that the wire is infinitely long is very important because without this we can not take vector E to be perpendicular to the curved part of the cylindrical gaussian surface.I think it should be parallel in place of perpendicular or it should be flat part rather than curved part.
No, it means perpendicular to the curved part, although I prefer to think of this as perpendicular to the axis of the cylinder i.e. the wire. This diagram and explanation works for me.
gracy said:
I also want to ask there is one more sentence that I am not getting
the expression of E derived from an infinitely long straight uniformly charged wire i.e Vector E= λ/2πε0r
is true largely at the central portion of the wire length and not really true at the ends of wire.I want to know why this is so?
An infinitely long wire doesn't have ends. This is important because it means that we can ignore what happens at the ends thus providing the symmetry which makes it inevitable that ## \vec E ## is perpendicular to the axis.
 
MrAnchovy said:
No, it means perpendicular to the curved part
But electric field vector is parallel to area vector of the curved part.
 
MrAnchovy said:
thus providing the symmetry which makes it inevitable that E⃗ \vec E is perpendicular to the axis.
how?
 
gracy said:
But electric field vector is parallel to area vector of the curved part.
Yes, that is because an area vector is perpendicular to the area to which it relates. The statement "the electric field vector is perpendicluar to [an area tangent to ] the curved part" is equivalent to "the electric field vector is parallel to the area vector of [an area tangent to ] the curved part".
 
gracy said:
how?
If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
 
MrAnchovy said:
If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
Except at midpoint,right?
 
gracy said:
Except at midpoint,right?
Exactly, the midpoint is the special case which necessitates the words "in general a point" rather than "every point".
 
MrAnchovy said:
thus providing the symmetry which makes it inevitable that E⃗ \vec E is perpendicular to the axis.
But how symmetry is relevant ?how does symmetry decide Whether E would be perpendicular or not?
 
  • #10
If and only if the cylinder is infinite, the influence of the charges tending to point E towards one end of the cylinder is equal to the influence of the charges tending to point E towards the other.
 
  • #11
MrAnchovy said:
If and only if the cylinder is infinite, the influence of the charges tending to point E towards one end of the cylinder is equal to the influence of the charges tending to point E towards the other.
Still not clear.
 
  • #12
MrAnchovy said:
the influence of the charges tending to point E towards one end of the cylinder is equal to the influence of the charges tending to point E towards the other.
Why are we talking about equaling up the effects of charges?I want to know how symmetry decides direction of vector E?
 
  • #13
gracy said:
Why are we talking about equaling up the effects of charges?I want to know how symmetry decides direction of vector E?
What other direction could it be? Assume that it pointed 80 degrees instead of perpendicular. How would you know of it is 80 degrees one way or 80 degrees the other way?
 
  • Like
Likes pbuk
  • #14
Only if the wire is infinitely long, the field will be a function of only the distance ("r" in cylindrical co-ordinates) from the wire,otherwise as the formula of Electric field at any point says it also depends on amount of source charge.
MrAnchovy said:
If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
This makes different amount of source charge to be at different distance .Thus making field different at most of the points.
But when we take the wire to be infinitely long, we can picture the field as being a function only of the distance ("r" in cylindrical co-ordinates) from the wire, the farther from the wire, the weaker it is. The loci of all points of any constant field strength we choose/select will be a cylinder with a central axis on the wire.
Right?
 
  • #15
gracy said:
I want to know how symmetry decides direction of vector E
Good approach ! Both lazy physicists and good physicists always keep their eyes open for symmetries :smile:.

The case cries out for using cylindrical coordinates ##\rho, \phi, z##

In this case there are at least four symmetries in evidence

1. rotational -- same situation when replacing ##\phi## by ##\phi + \alpha ## with a constant ##\alpha\quad \Rightarrow \quad ## no ## \phi ## dependence

2. translational -- same situation when replacing ##z## by ##z + p ## with a constant ## p \quad \Rightarrow \quad ## no ## z ## dependence

3. reflection -- same situation when replacing ##\phi## by ##- \phi ## in combination with #1 ## \Rightarrow \quad \phi ## component is zero

4. reflection -- same situation when replacing ##z## by ## -z ## in combination with #2 ##\Rightarrow \quad z ## component is zero

In short: Yes. (to your last question)
 
Last edited:
  • Like
Likes gracy
  • #16
BvU said:
(to your last question)
MY POST #14 "Right?'?
gracy said:
Right?
 
  • #17
Yes. really.
 
  • #18
BvU said:
Yes. really.
But how?
I thought I was wrong here
gracy said:
otherwise as the formula of Electric field at any point says it also depends on amount of source charge.
gracy said:
This makes different amount of source charge to be at different distance .Thus making field different at most of the points.
 
  • #19
I mean
pict.png
 
  • #20
My turn:
gracy said:
I also want to ask there is one more sentence that I am not getting
the expression of E derived from an infinitely long straight uniformly charged wire i.e Vector E= λ/2πε0r
is true largely at the central portion of the wire length and not really true at the ends of wire. I want to know why this is so?
What the book tries to bring across here is:

For a wire with finite length, the expression of E derived from an infinitely ...
which is probably also the context you left out in your post #1. Thus making the thread a confusion between infinitely long and finite length---

gracy said:

the answer is Yes. really.
 
  • Like
Likes gracy
  • #21
BvU said:
the answer is Yes. really.
How?
When the wire is of infinite length
G5man.png

when it has a finite length
YZ0Yl.png

I want to understand why this is so?
 
  • #22
Your picture doesn't really show a wire with a finite length. It shows a very thick, very short barrel. For normal folks to call something a wire, the length has to be considerably bigger than the diameter. You can't come close enough to your barrel to let the expression for the electric field approach the infinite wire expression.

The "infinite wire" approximation is valid as soon as you are so close to the wire that you can't see the ends -- or better: when the viewing angle for the whole wire approaches ##\pi/2##. Check out the worked out problem here : the exact expression is (11) and for L >> ##\rho## (his y) you get his last expression (15),
 
  • Like
Likes gracy
  • #23
gracy said:
Only if the wire is infinitely long, the field will be a function of only the distance ("r" in cylindrical co-ordinates) from the wire,otherwise as the formula of Electric field at any point says it also depends on amount of source charge.

This makes different amount of source charge to be at different distance .Thus making field different at most of the points.
But when we take the wire to be infinitely long, we can picture the field as being a function only of the distance ("r" in cylindrical co-ordinates) from the wire, the farther from the wire, the weaker it is. The loci of all points of any constant field strength we choose/select will be a cylinder with a central axis on the wire.
Right?
The electric field does depend on the amount of surface charge in any case.
The surface charge can be homogeneous along the finite wire or rod, still the electric field would change along the length. In case of infinite length, the parallel component of the charge elements cancel each other, only the normal component remains and the field lines are perpendicular to the wire .

finiterod.JPG
 
  • Like
Likes gracy
  • #24
Imagine the charge distributed evenly along the wire. Along the black line between the two charges, the resultant field is perpendicular to the wire as the parallel components cancel.
wireresfield.JPG
 
  • Like
Likes gracy
  • #25
ehild said:
Imagine the charge distributed evenly along the wire. Along the black line between the two charges, the resultant field is perpendicular to the wire as the parallel components cancel.
This is about infinite line,right?
 
  • #26
The lines which cancel each other are represented canceling each other in same color.
c.png

right?
 
  • #27
I KNOW it is silly question but Why the following is not correct?
ca.png
ca.png
 
Last edited:
  • #28
gracy said:
I want to know how symmetry decides direction of vector E?
I think I have got my answer
MrAnchovy said:
If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
gracy said:
Except at midpoint,right?
following is the situation in case of mid point
http://www.physics.sjsu.edu/becker/physics51/images/22_18EofLINE.JPG
But we want electric field to be equal at every points not just (in the line of)mid point and that too should be perpendicular to curved surface of cylinder (for ease in calculation )but there is a problem i.e
MrAnchovy said:
If the cylinder has finite length then in general a point on its surface is closer to one end than the other.
And as a result there would be some amount of electric field at any point on the curved surface in downward direction dEy along with perpendicular Electric fields lines dEx
And then net electric field will not remain perpendicular to the curved surface,because resultant of dEx And dEy simply can not be in direction of dEx .But when we take /assume length/axis /height of cylinder to be infinite
MrAnchovy said:
An infinitely long wire doesn't have ends.
so we can get symmetry .every dEy component (due to upper portion )would get canceled by the dEy component (due to bottom portion)as cylinder of infinite length does not have any length limits.Hence,we get net electric field perpendicular to curved surface.
Right?
 
Last edited by a moderator:
  • #29
gracy said:
How?
When the wire is of infinite length
@gracy please answer my questions in post 13.
 
  • #30
Isn't my post 28 correct?
 
  • #31
DaleSpam said:
please answer my questions in post 13.
I would not be able to.I did not understand the question.
 
  • #32
gracy said:
This is about infinite line,right?
Right.
 
  • #33
gracy said:
I would not be able to.I did not understand the question.
The point of the question is to get you to think about symmetry directly. For concreteness, let's assume a wire in a horizontal plane stretching off to infinity in both the left and right directions.

You are asking how the symmetry requires that the field be perpendicular to the wire. So, I am asking you the reverse. How could it possibly not be perpendicular? If it were not perpendicular then let's say that it points 80 degrees (just as an example) to the right. My question to you is, how would it know to point to the right instead of to the left?

Please think about it and answer the question directly. This is an exercise in thinking about symmetry.
 
  • #34
gracy said:
Isn't my post 28 correct?
Yes, your post in 28 is correct, but it also shows that you are thinking of it in terms of mechanisms rather than in terms of symmetry.
 
  • #35
DaleSpam said:
let's assume a wire in a horizontal plane stretching off to infinity in both the left and right directions.
I can also assume the wire to be along y-axis i.e vertical right?
 
  • #36
How would a wire be in a horizontal plane and along the vertical axis? No, the wire is horizontal, going left and right. So any vector which is perpendicular to the wire has no component either left or right.
 
  • #37
DaleSpam said:
How would a wire be in a horizontal plane and along the vertical axis?
I meant can I assume wire to be in vertical plane?
 
  • #38
"Up" and "down" are asymmetric, even though the asymmetry is not relevant for this problem, so you may unintentionally think of that as a reason for the field to point one way or the other. Let's stick with "left" and "right" which are clearly symmetric.
 
  • #39
Whew. So, getting back to basics. What IS an electric field or an E-vector? Well, its a mathematical construct which enables us to predict how things in the real world work. But consider measurement. One way to measure the field is measure the force at a "point". Ok, so we can measure magnitude and direction. For a wire (which unless you've good reasons not to, should be assumed to have "nearly zero" diameter - this is a convention for beginning Physics problems, not a "written in stone" definition) We can use x,y vector components. Its really a bad idea to talk about x,y and in the same thread vertical and horizontal. Make up your mind and stick to one or the other, don't waste time labeling them so that we can convert. A wire is 3 dimensional (in three dimensional space) but (see above) has negligible lateral cross-section. The field, of course will be symmetrical in planes perpendicular to the long axis...call it x-axis. This 'fact' suddenly reduces the important parameter to be r, the distance from the wire. Talking about surface effects with a wire means that you are no longer considering some "near zero radius" conductor of long length. So again, you have to make up your mind. At a 'sufficient' distance from said wire (said surface) the wire's effects can be considered as being from the center axis. Its not a bad question to ask "at what distance should we abandon the "near zero, no surface effects" paradigm. You can actually do some math and see how the assumption gets to be a worse and worse approximation as the distance, call it r/R° where R° is the radius of the wire, nears 1. A ratio of 10 will give 'not terrible' solutions but 100X would be better. So say at 50X, the force vectors break down to x,y,z components, but since (y,z) symmetry prevails, we just consider the distance, r, to the wire and can consider that the problem has been reduced to 2 dimensions, one of which is a "real" dimension (along the length of the wire) but the other dimension is a projection from the (yz) plane to a abstract dimension. In this scenario, for an infinite linear uniform charge distribution (a wire) every point charge to the left oh, sorry, to the minus x distance from any point will have an identical charge to the plus x distance from that point (don't have to call that point 0 or (0,0) or (0,0,0) could be ( 1, -3) in the reduced dimensions, or maybe (1, 3/√2,-3/√2) in real space (distance to x=0 is 3 in either case). This means for every force component → at any point (far enough from the wire) there is a ← component which cancels it from a charge at exactly the same distance in the opposite x direction. So,"by symmetry" ALL of these components cancel. Leaving only the r component of force. Hence the e field only 'depends' on this, "by symmetry". If you know something about antennas, you know that the end of a wire is where emr fields get nasty. Correctly terminating a wire requires care and is expensive in the real world. So, for beginning physics, we use infinite wires, no nasty terminal field effects, symmetrical, it makes problems so much easier. And can be a good approximation when assumptions are met. Note that lots of real world problems can't be 'reduced' by symmetry. Often these problems can only be solved by numerical (approximate) methods - there just aren't any simple solutions. In fact, most real world problems aren't "analytically tractable". There's no analytical solution to designing the "optimum" antenna, for instance.
 
Back
Top