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Finding capacitance

  1. Mar 30, 2014 #1
    I need guidance calculating the capacitance between two infinitely parallel long wires that are parallel to a large flat metal sheet.

    Condition:
    V = 0 at the origin.

    Both plane and lines have uniform charge.
    Left line: Negative charge density
    Right Line: Positive charge density

    Radius of lines:a
    Distance between lines: L
    Distane between lines and plane: D

    My attempt.
    Find Electric field (El) due to both lines at point P
    Find Electric field (Ep) due do plane at point P

    E at point P = EP = Ep + El

    ΔV(x,y) = -∫EPdy + C
    Now solve for C: C = ΔV(0,0) +∫EPdy|(0,0)

    Finding capacitance:

    C = Q/ΔV
    ΔV =
    ... not sure what to do from here...

    Is my approach correct so far?
     

    Attached Files:

  2. jcsd
  3. Mar 30, 2014 #2

    Simon Bridge

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    Start with neutral wires. The two wires start out at a non-zero potential (because of the charged sheet).

    Move charges a bit at a time from one wire to the other one.
    Initially this takes no work but gradually the work increases.

    What is the relationship between the work to move total charge Q and the capacitance.
     
  4. Mar 30, 2014 #3
    ΔV(x) = pL/(piε0)ln[(d-a)/a] = Change in potential from the surface of
    pL>0 to pL< 0 wire.
    pL: Units C/m

    i know you asked for the work but is it ok if i just calculate the potential?
     
    Last edited: Mar 30, 2014
  5. Mar 30, 2014 #4

    Simon Bridge

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    The idea is to use some physics to work out the capacitance.
    The work moving the charge is the energy stored in the capacitor ... which has a nice relationship to the capacitance and is general.

    But the suggestion is mostly just to nudge you into thinking in terms of physics, so do it whichever way you feel most comfortable.
     
  6. Mar 30, 2014 #5
    Ok well the work in moving charge q is
    q(pL/(piε0)ln[(d-a)/a])

    The total work in moving charge q from the right wire to the left then to the plate is...

    q{pL/(piε0)ln[(d-a)/a] + psd/(2ε0)} = qΔV

    pL = QL/L
    ps = Qs/A ; A = L2
     
    Last edited: Mar 30, 2014
  7. Mar 31, 2014 #6
    Am I doing your approach correctly?
    Actually the work in the y direction should be due to boh the line charges and plate... I didn't include the line charges.
     
  8. Mar 31, 2014 #7

    Simon Bridge

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  9. Mar 31, 2014 #8
    Sorry, I left all the integration out..
    This is what I did.
    General Eq. for Eline = pL/(2piε0r),
    1.r is the distance from the line and
    2. pL is the charge density of the line

    The given the problem.
    Eline = ELeft Line + ERight Line
    = pL/(2piε0)[1/(d/2 - x) + 1/(d/2+x)] { points in the neg, x direction}

    d/2 - x is the distance from an arbitrary point (x,0) to the right line
    d/2+x is the distance from an arbitrary point (x,0) to the left line

    Now.. V(x) = -∫Edx = pL/(2piε0)ln[(d/2+x)/(d/2-x)] + C
    and
    V(0) = 0 = C.
    Now the change in potential from the right line to the left line is.
    ΔV = pL/(2piε0)ln[(d-a)/a]
     
    Last edited: Mar 31, 2014
  10. Mar 31, 2014 #9

    Simon Bridge

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    Without checking your arithmetic - OK.
    So where do you go from there?
     
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