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Gauss' Law --> Electric field of an infinitely long, straight line

  1. Aug 10, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Problem statement:
    An infinitely long, straight line has a uniform charge distribution of ρ C/m. Use Gauss' law to find the electric field at a point r m away from it.

    Solution:
    Consider a cylindrical volume of height ℓ with circular cross sectional area of radius r, which has the line as its axis. The volume contains a total charge of Q = ρℓ. By symmetry, the E field is radial in direction and has the same magnitude on the surface of the cylinder. The total flux through the surface is Ψ = ϵE × ℓ × πr^2. By Gauss' law, Ψ = Q from which E = ρ/(πϵr^2).

    2. Relevant equations
    Gauss' Law.
    Cylinder.

    3. The attempt at a solution
    I'm trying to understand the given solution.

    Is Q = ρℓ instead of Q = ρV true, because the charge is distributed along the real-world line instead of the imaginary cylinder?

    And, what about Gauss' Law indicates that Ψ = Q?

    Also, I keep seeing the capital letter phi, Φ, in the context of Gauss' Law. Is Φ from what I see on-line the same thing as the Ψ I see in my book?

    Lastly, how was the total flux found to be Ψ = ϵE × ℓ × πr^2?

    For what it's worth, given Ψ = Q, Q = ρℓ and Ψ = ϵE × ℓ × πr^2, I do know how to get E = ρ/(πϵr^2).

    Any input would be GREATLY appreciated!
     
  2. jcsd
  3. Aug 10, 2014 #2

    rude man

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    Yes. In either case it's the free charge within the volume πr2ℓ.
    There is no standard symbol for the electric E or the displacement D flux. Most of the time the flux is considered as the integral over the closed surface of the E field. But Gauss says ∫∫D*dA = Qfree. Apparently your book likes to call ψ = ∫∫D*dA so then ψ = Qfree.

    It is necessary to use free charge only in the above expression for ψ; bound charge in a dielectric doesn't count. D includes the effect of dielectric constant; E does not.

    D = εE so the total D flux ψ = ∫∫D*dA = ϵE × ℓ × πr^2.

    Vectors in bold; "*" signifies dot product.
     
  4. Aug 12, 2014 #3
    Before I begin, I would like to remind you that Gauss' law is:
    $$\frac{Q}{\epsilon_0} = \oint \vec E \cdot d\vec A $$

    You have an infinitely long wire. Your wire is cylindrical I assume, and so there will be a symmetrically radiating e-field.

    How do you take advantage of this with Gauss law? Well, if we pick a Gaussian surface that encloses the e-field so that the e-field is the same all around the surface, than we can take the electric field out of the surface integral in Gauss law, since the e-field will be a constant.

    What shape encapsulates a cylindrically radiating e-field so that the e-field is the same everywhere on the surface? A cylinder of course, so you would draw your cylindrical Gaussian surface around the wire.

    The total flux through your cylinder will be:
    $$\frac{Q}{\epsilon_0} = \oint_{side} \vec E \cdot d\vec A + \oint_{base}\vec E \cdot D\vec A$$

    Since the base of the cylinder (the circle) is always perpendicular to the radial electric field from the wire, the cosine in the dot product between the two will be zero, and so you can take the surface integral for the base out of the equation:
    $$\frac{Q}{\epsilon_0} = \oint_{side} \vec E \cdot d\vec A $$

    E and dA are always parallel to each other so the cosine in the dot product between the two is one, thus we can simplify the equation to be:
    $$\frac{Q}{\epsilon_0} = \oint_{side} E \cdot dA $$

    We know that the electric field intensity everywhere on the side of the cylinder will be the same by the symmetry of the cylinder and the radial electric field; as a result, we can take E out of the surface integral.

    $$\frac{Q}{\epsilon_0} = E \oint_{side}dA $$
    The surface integral of small side areas dA for the side of the cylinder is just the area of the side of the cylinder itself:
    $$\frac{Q}{\epsilon_0} = E \cdot A_{side} $$

    Solve for the E-field of the wire:

    $$E = \frac{Q}{\epsilon_0 \cdot A} = \frac{Q}{\epsilon_0 \cdot 2\pi r l}$$

    The charge distribution represents the charge per unit of length, so the the amount of charge enclosed in the Gaussian surface (that encapsulates the long wire/line) is just the charge distribution per unit of length multiplied by the length. Thus, you can replace Q with:
    $$E = \frac{Q}{\epsilon_0 \cdot A} = \frac{\rho \cdot l}{\epsilon_0 \cdot 2\pi r l}$$
    The two l's cancel, and you are left with the answer for the e-field:
    $$E = \frac{Q}{\epsilon_0 \cdot A} = \frac{\rho}{\epsilon_0 \cdot 2\pi r }$$

    By the way, I am not sure how you got that the solution was
    ρ/(πϵr^2).
     
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4

    rude man

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    I missed that. It's incorrect. You used the area of one end of the gaussian cylinder and used ρ instead of Q in the numerator. The area is 2πrl so E = Q/2πεrl = ρ/2πεr. On the other hand, you were correct in using ε instead of ε0. The problem did not specify air or a vacuum as the dielectric, it could have been a solid, fluid or gas with ε > ε0. Even water has ε = 80.4ε0!
     
  6. Aug 31, 2014 #5

    s3a

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    Thank you for your (very useful :)) responses, and sorry for my delayed response.

    So, just to be clear, that means that Gauss's Law should generally state that ∮(D ⋅ dA) = Q/ϵ instead of ∮(E ⋅ dA) = Q/ϵ_0, and that it should also generally be the case that D = ϵE + P instead of D = ϵ_0 E + P (where P = 0 in the case of the problem of this thread, and where ϵ may or may not equal to ϵ_0)?

    Assuming I am right, why do I keep seeing ϵ_0 instead of ϵ in books and Wikipedia? Is it simply because "everyone" is assuming that a vacuum is being considered?

    P.S.
    I now know it's wrong (thanks to wenqin123), but, for what it's worth, I got E = ρ/(πϵr^2) by working with the following three equations.:

    (1) Ψ = Q

    (2) Q = ρℓ

    (3) Ψ = ϵE × ℓ × πr^2


    I used the above three equations as follows.:

    Ψ = Q and Q = ρℓ ⇒ Ψ = ρℓ

    as well as

    Ψ = ρℓ and Ψ = ϵE × ℓ × πr^2 ⇒ ρℓ = ϵE × ℓ × πr^2 ⇒ ρ = ϵE × πr^2 ⇒ E = ρ / (ϵ × πr^2).
     
  7. Aug 31, 2014 #6

    s3a

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    rude man, I just realized this was mentioned in your first post.

    However, when you say ψ = ∫∫D*dA, do you mean ∮(D ⋅ dA) = ∮([Eϵ + P] ⋅ dA) = ∮([Eϵ + 0] ⋅ dA) = ∮([Eϵ] ⋅ dA) = Q ⇒ ∮(E ⋅ dA) = Q / ϵ?

    P.S.
    Notice I used "ϵ" instead of "ϵ_0".

    Edit:
    Please make sure you read the latest/edited version of this post.
     
  8. Aug 31, 2014 #7

    rude man

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    First, I would reserve the single integral sign with the circle in it for contour, not surface, integration.

    Second: You sort of went from wrong to right. The corrected sequence would be

    ∫∫D.dA = ∫∫(ε0E + P).dA = ∫∫εE.dA = Qfree ⇒ ∫∫E.dA = Qfree/ε.

    I emphasize Qfree because bound charges, as in a dielectric, are excluded.
     
    Last edited: Aug 31, 2014
  9. Sep 1, 2014 #8

    s3a

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    Alright, thank you both, I think I have what I need. :)
     
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