# Doubts about Electric field due to an infinitely long wire

1. Sep 13, 2015

### gracy

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?

2. Sep 13, 2015

### MrAnchovy

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.
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.

3. Sep 13, 2015

### gracy

But electric field vector is parallel to area vector of the curved part.

4. Sep 13, 2015

### gracy

how?

5. Sep 13, 2015

### MrAnchovy

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".

6. Sep 13, 2015

### MrAnchovy

If the cylinder has finite length then in general a point on its surface is closer to one end than the other.

7. Sep 13, 2015

### gracy

Except at midpoint,right?

8. Sep 13, 2015

### MrAnchovy

Exactly, the midpoint is the special case which necessitates the words "in general a point" rather than "every point".

9. Sep 13, 2015

### gracy

But how symmetry is relevant ?how does symmetry decide Whether E would be perpendicular or not?

10. Sep 13, 2015

### MrAnchovy

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. Sep 13, 2015

### gracy

Still not clear.

12. Sep 13, 2015

### gracy

Why are we talking about equaling up the effects of charges?I want to know how symmetry decides direction of vector E?

13. Sep 13, 2015

### Staff: Mentor

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?

14. Sep 14, 2015

### gracy

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?

15. Sep 14, 2015

### BvU

Good approach ! Both lazy physicists and good physicists always keep their eyes open for symmetries .

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: Sep 14, 2015
16. Sep 14, 2015

### gracy

MY POST #14 "Right?'?

17. Sep 14, 2015

### BvU

Yes. really.

18. Sep 14, 2015

### gracy

But how?
I thought I was wrong here

19. Sep 14, 2015

### gracy

I mean

20. Sep 14, 2015

### BvU

My turn:
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

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