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Capacitance of eccentrical cylinders

  1. Jul 19, 2004 #1
    Hi, I have a problem about the capacitance of two cylinders eccentrically located one inside the other - with radii a and b resp., their centers have a distance c apart. I've tried that with the method of images, considering the eqipotential "cylinders" of a system of two infinite parallel lines of charges at distance 2d apart.

    The potential at any point P on the plane perpendicular to the lines is given by V = (q/(2Pi*Eps))ln(R2/R1)
    where q is the charge density
    R1 and R2 are the distances from line 1 and line 2 respectively, i.e. r1^2 = (x-d)^2+y^2 and r2^2 = (x+d)^2+y^2

    To get eqipotential lines of the system, we equate R2/R1 to a constant k and then we get a family of circles of the plane with center at h = d(k^2+1)/(k^2-1) and radius R = [2dk/(k^2-1)] ^2

    Everything seems fine. If i can figure out the two values of K corresponding to the two equipotential "circles", I will know the p.d, as well as the capacitance.

    However, I have problem to solve them out without knowing the value of d.

    Can someone help? :smile:

    Thanks very much.


    Carl T.A. Johnk, Engineering Electromagnetic Fields and Waves 2nd Ed., pp. 222-225
  2. jcsd
  3. Jul 20, 2004 #2


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    I'm sorry that no one seems to want to help you with this one. I will give it a shot.

    I do not understand what you are trying to do with R2/R1 = k. You already have the equipotential surfaces given by:

    V = (λ/2π&epslion;0) ln(ρ21).

    One of the tricks is to realize that 2d = ρ1 + ρ2.

    The other basic trick is to realize that the distance between the plates is b - (a + c), and that this distance represents the distance between the two different equipotential surfaces of interest.

    Finally, a word of caution: remember that, in general, neither image charge will lie at the center of neither conductor.

    I hope I didn't just tell you stuff that you already know.
  4. Jul 20, 2004 #3


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    The answer you want for the capacitance per length is

    C= 2pi*epsilon*{1/[inverse cosh(-1*(c**2-a**2-b**2)/2ab)]}

    the units are such that the Coloumb potential is (1/4pi*epsilon)Q/R

    This is worked out in the book Static and Dynamic Electricity by Symthe, Chapter 4. --once the scourge of most physics graduate students, among others, we prayed that such a problem would not turn up in our qualifiers or orals. You are posing a difficult problem, which is solved via complex variable approaches also used in fluid dynamics, stream functions and all that. Symthe wrote his book in the late 1930s at which time the emphasis on mathematical detail and prowess was considerable different than it is now.

    reilly Atkinson
  5. Jul 20, 2004 #4


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    That looks so familiar. What does the ** (double asterisk) mean?
  6. Jul 20, 2004 #5


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    Turin -- a**2 means a*a, harks back to FORTRAN. in which ** means exponentiation Regards, Reilly
  7. Jul 20, 2004 #6
    Thanks reily and turin,
    It turned out that I arrived at the same answer with the help of Mathematica without bothering the complex analysis technique.
    turin, why i care about the constant k? Coz it gives me the family of circles that represent the equipotential lines. So each of the two conducting cylinders should be one in the family with different k's with center at (h,0) and radius R.
    [tex]h = d \frac{k^2+1}{k^2-1} (1)[/tex]
    [tex]R = \frac{2dk}{k^2-1} (2)[/tex]

    Let say the center of the inner one is at (h1, 0), and the other is at (h2,0).
    Then [tex]h1 = d \frac{k1^2+1}{k1^2-1})[/tex] and [tex] h2 = d \frac{k2^2+1}{k2^2-1})[/tex] by (1).

    Their difference c turns out to be simple becoz of (2).
    [tex]c = \frac{r2}{k2} - \frac{r1}{k1} (3) [/tex]
    The ratio of the radii gives
    [tex]\frac{r1}{r2} = \frac{k1}{k2} \frac{k2^2-1}{k1^2-1} (4) [/tex]
    Having (3) and (4), I just let Mathematica do the rest to solve k1 and k2 for me.

    Finally, the voltage difference between the two follows [tex] \frac{q}{2\pi \epsilon L} \ln \frac{k1}{k2} [/tex]
    The capacitance per length agrees with Symthe's, which is [tex] \frac{2\pi \epsilon}{arccosh (\frac{a^2+b^2-c^2}{2ab})} [/tex]

    woo... beautiful formula. Thx folks.

    Last edited: Jul 20, 2004
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