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A tricky problem on capacity

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Two identical conducting spheres of radii "a" are kept in contact. Find the capacity of the system

    2. Relevant equations

    C=Q/V

    3. The attempt at a solution

    Since the two spheres are kept in contact so they are at the same potential.So we need to find out this potential so as to find the capacitance. But I am not able to start at the solution
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 13, 2010 #2
    How far are they?
     
  4. Aug 13, 2010 #3
    They are kept in contact
     
  5. Aug 13, 2010 #4
    So they are placed next to each other, not through any wire?
     
  6. Aug 13, 2010 #5
    No, directly in contact
     
  7. Aug 13, 2010 #6
    Capacitors works creating an electric field across a dielectric, I'm pretty sure that if the conducting spheres are in touch the capacitance would be zero.
     
  8. Aug 13, 2010 #7
    This is a very hard problem.

    EDIT:

    Let's take the z-axis along the line connecting the two centers of the two spheres, and the point where the spheres touch eachother lie in the xy-plane (it follows that this point coincides with the origin).

    From the geometry of the problem, it follows then that the z-axis is an axis of symmetry and that the xy-plane is a plane of symmetry for the problem. Therefore, the potential [itex]\Phi[/itex] does not depend on the azimuthal angle [itex]\phi[/itex] and is also an even function of [itex]z[/itex]. It also satisfies the Laplace equation [itex]\nabla^{2} \, \Phi = 0[/itex] everywhere outside the spheres with the boundary conditions that [itex]\Phi \rightarrow 0[/itex] at infinity and [itex]\Phi = \Phi_{0}[/itex] on the spheres.

    Let us try and solve this boundary value problem. Before we go any further, let us try and "invent" a new coordinate system, where one set of coordinate surfaces are spheres with the center on the z-axis and such that the origin always lies on the sphere. If the sphere has radius [itex]a[/itex], then, the equation for such a sphere in cylindrical coordinates is:

    [tex]
    \rho^{2} + (z - a)^{2} = a^{2}
    [/tex]

    [tex]
    \rho^{2} + z^{2} - 2 a z = 0
    [/tex]

    The differential equation for this family of surfaces is obtained by eliminating the parameter [itex]a[/itex]:

    [tex]
    2 \rho + 2 z z' - 2 a z' = 0 \Rightarrow 2 z (\rho + z z') - 2 a z z' = 0
    [/tex]

    [tex]
    z' (\rho^{2} + z^{2}) - 2 z (\rho + z z') = 0
    [/tex]

    [tex]
    (\rho^{2} - z^{2}) z' - 2 \rho z = 0
    [/tex]

    A family of orthogonal surfaces is obtaned by the substitution [itex]z \rightarrow -1/z[/itex]:

    [tex]
    (\rho^{2} - z^{2}) (-\frac{1}{z'}) - 2 \rho z = 0
    [/tex]

    [tex]
    \rho^{2} - z^{2} + 2 \rho z z' = 0
    [/tex]

    [tex]
    z' = \frac{z^{2} - \rho^{2}}{2 \rho z}
    [/tex]

    This is a homogeneous differential equation and the variables can be separated by applying the substitution [itex]u = z/\rho \Rightarrow z = \rho u, z' = u + \rho u'[/itex]:

    [tex]
    \rho u' + u = \frac{u^{2} - 1}{2 u}
    [/tex]

    [tex]
    \rho u' = - \frac{u^{2} + 1}{2 u}
    [/tex]

    [tex]
    \frac{2 u du}{u^{2} + 1} + \frac{d\rho}{\rho} = 0
    [/tex]

    [tex]
    \ln{(u^{2} + 1)} + \ln{\rho} = c'
    [/tex]

    [tex]
    \rho (u^{2} + 1) = C
    [/tex]

    [tex]
    \rho^{2} + z^{2} - C \rho = 0
    [/tex]

    But, this determines a sphere of radius [itex]|b|, \; b = C/2[/itex] lying on the [itex]\rho[/itex]-axis and touching the origin.

    Let us use a and b as new coordinates instead of [itex]\rho[/itex] and z. We need to solve the system of equations:

    [tex]
    \left\{\begin{array}{lcl}
    \rho^{2} + z^{2} - 2 a z & = & 0 \\

    \rho^{2} + z^{2} - 2 b \rho & = & 0
    \end{array}\right. \Rightarrow \left\{\begin{array}{l}
    \rho = \frac{2 a^{2} b}{a^{2} + b^{2}} \\

    z = \frac{2 a b^{2}}{a^{2} + b^{2}}
    \end{array}\right. \Rightarrow r^{2} = \rho^{2} + z^{2} = \frac{4 a^{2} b^{2}}{a^{2} + b^{2}}
    [/tex]

    With these definitions, the metric becomes:

    [tex]
    ds^{2} = d\rho^{2} + dz^{2} + \rho^{2} d\phi^{2} = \frac{4(b^{4} \, da^{2} + a^{4} \, db^{2}}{(a^{2} + b^{2})^{2}} + \frac{4 a^{4} \, b^{2}}{(a^{2} + b^{2})^{2}} d\phi^{2}
    [/tex]

    and the Laplace operator takes the form:

    [tex]
    \nabla^{2} \Phi = \frac{
    (a^{2} + b^{2})^{3}}{4 b^{4}} \, \frac{\partial}{\partial a}\left(\frac{a^{4}}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial a}\right) + \frac{(a^{2} + b^{2})^{3}}{4 a^{2} \, b^{3}} \, \frac{\partial}{\partial b}\left(\frac{b}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial b}\right) + \frac{(a^{2} + b^{2})^{2}}{4 a^{4} b^{2}} \, \frac{\partial^{2} \Phi}{\partial \phi^{2}} = 0
    [/tex]

    Assuming azimuthal symmetry, this potential does not depend on [itex]\phi[/itex]. However, the equation does not allow separation of the variables a and b.
     
    Last edited: Aug 13, 2010
  9. Aug 13, 2010 #8
    when the two spheres are kept in contact, the capacitance of the system of two spheres will be zero..
    but mind you, its not that what you are being asked. Its capacity of the spheres!!
    for example, when we are asked to determine the capacity of a sphere carrying a charge, we assume the other concentric sphere is at infinity with opposite charge, so we can find the expression for electric field and hence integrate E.ds with apt limit to find the potential difference and hence the capacity of the sphere,
    the question here is how can you determine one unique expression of electric field in this case at any point in space???
     
  10. Aug 13, 2010 #9
    edit: i was wrong
     
    Last edited: Aug 13, 2010
  11. Aug 13, 2010 #10

    gabbagabbahey

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    This is misleading, a single conducting spherical shell does not qualify as a capacitor. Moreover, the [itex]V[/itex] and [itex]Q[/itex] in the formula [itex]C=\frac{Q}{V}[/itex] are defined as the potential difference between the two plates of the capacitor and the charge on the positive plate respectively (the other plate must have a charge of [itex]-Q[/itex]).

    This is just wrong. According to your calculations, the Capacitance of two idendical conductiing spheres is the same no matter how far apart they are. Capacitance is a purely geometrical quantity, and should definately depend on how far apart the two plates (spheres in this case) are.
     
    Last edited: Aug 13, 2010
  12. Aug 13, 2010 #11

    gabbagabbahey

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    Haven't you already been warned about providing too much help for homework problems? In any case, you misinterpreted the problem.
     
  13. Aug 13, 2010 #12

    gabbagabbahey

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    If two conductors (any sphape and size) are in contact, what must be the potential difference between them (and why?:wink:)? What does that make [itex]V[/itex] in the capacitance formula [itex]C=\frac{Q}{V}[/itex]?
     
  14. Aug 13, 2010 #13
    Don't forget that even a single metal has capacitance albeit very small.A simple way to solve this problem would be to consider the two spheres being separated,one of them given a charge and the two then joined together.This is how I see the problem and I think it works.
     
  15. Aug 13, 2010 #14

    vela

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    Actually, it does.

    http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

    When I was taking E&M as an undergrad and first heard the question "What's the capacitance of a sphere?" I remember thinking, "That doesn't make sense. Where's the other conductor?" It turned out the question was about this notion of self-capacitance.
     
  16. Aug 13, 2010 #15

    vela

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    This problem suggests to me using a conformal mapping. Is that something you've covered? Unfortunately, I don't remember how to do this type of problem anymore, but I do remember that with the right mapping, the problem typically becomes very easy to solve.

    I found a good list of conformal mappings at

    http://math.fullerton.edu/mathews/c2003/ConformalMapDictionary.1.html [Broken]

    Map 13, in particular, might be applicable.
     
    Last edited by a moderator: May 4, 2017
  17. Aug 13, 2010 #16

    gabbagabbahey

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    Thanks, I don't remember ever studying self-capacitance... with this concept it seems as though the other conductor is a theoretical spherical shell of infinite radius. This seems like a useless concept to me... whenever you add charge to a conductor, it has to come from somewhere else (charge conservation). To me, that somewhere else is always the other half of the capacitor, not some theoretical infinite radius spherical shell. I guess there must be some practical use for this concept though?
     
  18. Aug 13, 2010 #17
    When we find the potential of a metal(V=Q/4pi* epsilon* r)for a sphere of radius r,V stands for the potential with reference to a zero potential in other words it stands for potential difference and the other conductor can be considered to be the earth.
     
  19. Aug 13, 2010 #18
    I can't think of any uses but would be interested to hear if there are any.The capacitance is incredibly small being just a small fraction of a Farad for a sphere as big as the earth.
     
  20. Aug 13, 2010 #19

    gabbagabbahey

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    Anyways, it definitely sounds like the problem is to calculate the self-capacitance of the two-sphere conductor. Since the problem has azimuthal symmetry (if you set up your Coordinate system so that the spheres are described by [itex]x^2+y^2+(z\pm a)^2=a^2[/itex] ), I'd use separation of variables in Cylindrical Coordiantes to solve Laplace's equation. Your boundary conditions should be obvious: the potential at infinity is zero, and the spheres are at some constant potential [itex]V_0[/itex]. Finding the potential everywhere will allow you to calculate the charge density on the conductor, and the total charge as a function of [itex]V_0[/itex]...divide that charge by [itex]V_0[/itex], and you'll be left with the self-capacitance.
     
  21. Aug 13, 2010 #20

    vela

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    I suppose it might be useful to model a system where charge collects on a conductor. That section from Wikipedia mentioned two applications: a van de Graaff generator and the Earth.

    I think it's used primarily to torture poor students like sagardip. Of course, classical electromagnetics is one of those subject that been around too long in that people have had time to solve all sorts of weird problems that now get assigned as homework. :wink:
     
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