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Potential of a long charged wire

  1. Jun 7, 2010 #1
    I'm studying electromagnetism through a book of Purcell and I'm having trouble in some point...

    The book tells me that I can calculate the potential of any charge distribution by this equation:

    [tex]\varphi(x; y; z) = \int_{all \ sources}^{ } \frac{\rho(x'; y'; z')dx' \ dy' \ dz'}{4\pi \varepsilon_{0}r } [/tex]

    So, the book says more things about this equation and I don't understand some things...

    [PLAIN]http://img717.imageshack.us/img717/8237/electricimagnetism1.jpg [Broken]

    1. Why the integration above won't work unless all sources are confined to some finite region of space? I understand it won't work if the charges are spread in all dimensions of the space... but if it's spread only in one dimension (like a infinitely long charged wire) the integration will work normaly (you just need to take the reference point as a point which has an infinite distance from the wire, r = +inf), isn't it?

    2. What does he mean with "the integral diverges"? Why do we get an infinite result?

    3. He says that we haven't this difficulty (charges not confined to an finite region of space) in calculating the electric field of the infinitely long wire because the contributions of elements of the line charge to the field decrease so rapidly with distance. Why? What the "decrease speed" of the electric fields has to do with the possibility to calculate the electric field of charges that are not confined to a finite region of space?

    4. He says that it's better to locate the zero of potential somewhere close to the wire. Why can't we still considering the zero of potential to a point that has r = +inf from the wire?

    I would be grateful if someone help me to proceed answering me these questions...

    Thank you,
    Rafael Andreatta
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 7, 2010 #2
    If you set the reference point at infinity you end up calculating ln(infinity) which diverges. Since the system contains an infinite amount of charge it´s not surprising the integral diverges when we move towards infinite. The same happens with uniformly charged plane.
     
  4. Jun 7, 2010 #3
    Thank you for the reply.

    Why does ln(infinity) diverges? What you mean by "diverge"? (I think it's a math question, but I won't open a thread on math forum just for that, could someone answer me?)
     
  5. Jun 7, 2010 #4

    rcgldr

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    It diverges because ln(∞) = ∞. Since reference points are arbitrary, there's no reason not to choose 1 unit of distance as the reference point for a long wire, since ln(1) = 0. For an infinte disc or plane the field strength is constant, so the reference point is normally at the surface of the plane (zero), similar to gravitational potential near the surface of the earth being defined as V = g h.

    Here is a link also showing the math for a infinite wire or cylinder, which continues on to handle the the field from a disc.

    http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elelin.html#c1

    The math for field strength from a solid disc is explained later on that same web page:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elelin.html#c3

    as [tex]R^2 \ \rightarrow \infty [/tex]

    then [tex] \frac{z} {\sqrt{z^2 + R^2}} \ \rightarrow \ 0 [/tex]

    and you end up with [tex]E_z = k \ \sigma \ 2 \ \pi [/tex]

    An alterative approach is to consider the field from an infintely long line (= 1/z), then integrate an infinitely large plane composed of infintely long rectangles that approach infinitely long lines as their width approaches zero.

    For an infinite line charge, E = 2 k λ / r, where λ is charge per unit length and r is distance from the line. Here is the derivation:
    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c1 [Broken]

    For an infinite plane case, infinitely long strips (rectangles) approximate a line as their width -> 0. The area of a strip of lengh L is L dx, and the charge dq = σ L dx. The charge per unit length dλ = dq/L = σ dx. Assume the plane exists on the x-y plane, then the magnitude of the field at any point in space from a strip is dE = 2 k σ / r, where r is the distance from a strip to that point in space. In the case of the entire plane, the x components cancel because of symmetry, with only a net force in the z direction, and for each strip of the plane, dEz = dE (z / r) = dE sin(θ).

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{{sin}(\theta)dx}{r}[/tex]

    [tex]{sin}(\theta) = z / r[/tex]

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{z {dx}}{r^2}[/tex]

    [tex]r^2 = z^2 + x^2[/tex]

    [tex]E = 2 k \sigma z \int_{-\infty}^{+\infty} \frac{dx}{z^2 + x^2}[/tex]

    [tex]E = 2 k \sigma z \left [ \frac{1}{z} tan^{-1}\left (\frac{x}{z}\right )\right ]_{-\infty}^{+\infty} [/tex]

    [tex]E = 2 k \sigma z \left ( \frac{1}{z} \right) \left ( \frac{+ \pi}{2} - \frac{- \pi}{2} \right ) [/tex]

    [tex]E = 2 \pi k \sigma [/tex]
     
    Last edited by a moderator: May 4, 2017
  6. Jun 8, 2010 #5
    Thank you for the reply.

    But I still don't understand some things: see my questions 1 and 3

    Thank you very much...
     
    Last edited by a moderator: May 4, 2017
  7. Jun 8, 2010 #6

    rcgldr

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    1. Finite distance, as opposed to infinte.
    This is because potential from an infinite line is relative to ln(r) (r is distance from the line), and ln(∞) = ∞.

    3. The field isn't an issue because it's relative to 1/r and 1/∞ = 0.
     
  8. Jun 8, 2010 #7
    Humm, now I got it, thank you very much.
     
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