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Capicatance of spherical capacitor

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A spherical capacitor consists of a spherical conducting sheell of radius b and charge -Q concentricc with a smaller conducting sphere of radius a and charge +Q. The capacitance of this device :

    a. zero

    b. [tex]\frac{ab}{4\pi\epsilon_0(b-a)}[/tex]

    c. [tex]\frac{ab}{4\pi\epsilon_0(a-b)}[/tex]

    d. [tex]\frac{4\pi\epsilon_0 ab}{b-a}[/tex]

    e. [tex]\frac{4\pi\epsilon_0 ab}{a-b}[/tex]


    2. Relevant equations
    Q = CV

    [tex]V=k\frac{Q}{r}[/tex]

    3. The attempt at a solution

    [tex]V_1=k\frac{-Q}{b}[/tex]

    [tex]V_2=k\frac{Q}{a}[/tex]

    Then, how to continue?

    Thanks
     
  2. jcsd
  3. Jan 18, 2010 #2

    rl.bhat

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    Re: Capacitor

    By applying Gauss' law to an charged conducting sphere, the electric field outside it is found to be
    E = kQ*r^2.
    The potential difference delta V = (kQ/r^2)*dr

    The voltage between the spheres can be found by integrating the electric field along a radial line:
    V =kQ*intg(1/r^2)*dr between a to b.
    From the definition of capacitance, the capacitance is
    C = Q/V.
     
  4. Jan 18, 2010 #3
    Re: Capacitor

    Hi rl.bhat

    V =kQ*intg(1/r^2)*dr between a to b = kQ (1/a - 1/b). This will be the same as V2 + V1 from my previous work.

    Should the potential difference be : V2 - V1 = kQ (1/a + 1/b)?

    Thanks
     
  5. Jan 18, 2010 #4

    rl.bhat

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    Re: Capacitor

    Inside the sphere the electric field is due to the smaller sphere only. The potential of the inner sphere with respect to the outer sphere is
    Vab = Va - Vb = kQ (1/a - 1/b).
     
  6. Jan 18, 2010 #5
    Re: Capacitor

    Hi rl.bhat

    I think I get your point. Now I have new questions.

    1.
    What if the charges are reversed (the outer sphere is Q and inner sphere is -Q)? Will it be :
    V =k(-Q)*intg(1/r^2)*dr between a to b

    2.
    What if the inner and outer spheres have the same charge, e.g -Q ?


    Thanks
     
  7. Jan 18, 2010 #6

    rl.bhat

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    Re: Capacitor

    Capacitance of the capacitor does not depends on the charges of the conductors. While performing its work it must have opposite charges on them.
    Capacitance of capacitor in the given problem depends on which conductor is grounded.
     
  8. Jan 19, 2010 #7
    Re: Capacitor

    so if the charges are reversed (the outer sphere is Q and inner sphere is -Q), it will be the same : V =kQ*intg(1/r^2)*dr between a to b ?

    then if the inner and outer spheres have the same charge, it is not capacitor, it is just conducting sphere with charge?

    In the question, the conductor grounded is the outer sphere?

    Thanks
     
  9. Jan 20, 2010 #8

    rl.bhat

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    Re: Capacitor

    Yes.


    [/QUOTE]then if the inner and outer spheres have the same charge, it is not capacitor, it is just conducting sphere with charge?[/QUOTE]

    It is not possible to have the same charge on the concentric spherical conductors out of which one is grounded.

    [/QUOTE]In the question, the conductor grounded is the outer sphere?[/QUOTE]

    In that case the outer sphere cannot be charged by any external source.
    Even you can ground the inner surface of the inner sphere instead of outer surface of the outer sphere.
     
  10. Jan 20, 2010 #9
    Re: Capacitor

    I don't read any infomation from the question about grounding. The outer sphere is more negative, so does it mean that the outer sphere is grounded?

    If I ground the inner sphere, will it be more negative than the outer sphere?

    Thanks
     
  11. Jan 20, 2010 #10

    rl.bhat

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    Re: Capacitor

    If you give a positive charge to the inner sphere, a negative charge will induce on the inner surface of the outer sphere and a positive charge on the outer surface of the outer sphere. Since in the problem, it is mentioned that outer sphere has negative charge, it is obvious that its outer surface is grounded. Negative charge on the inner remain there because they are bounded by the positive charge on the inner sphere.
    Grounding does not increase the charge on any sphere.
    If you want to ground the inner sphere, you have to ground the inner surface of the inner sphere. In that case, the system becomes the parallel combination of two capacitors
     
  12. Jan 20, 2010 #11
    Re: Capacitor

    Hi rl.bhat

    What will happen if I ground the outer surface of the inner sphere and why the system become parallel combination if I ground the inner surface of inner sphere?

    Sorry for asking so many questions. Thanks
     
  13. Jan 20, 2010 #12

    rl.bhat

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    Re: Capacitor

    Then you will have a single spherical capacitor of larger radius. Inner sphere is irrelevant. Charge will be stored only on the outer surface of the outer sphere.
    If you ground the inner surface of inner sphere, charge can be stored on inner and outer surfaces of the outer sphere.
    The combined capacitance will be
    C = 4π*εο*ab/(b - a) + 4π*εο*b
    Usually the capacitors are charged by removing the electrons from the conductors by any method. When you remove electrons from the outer surface of the outer sphere, it becomes positive. Electrons from the inner surface will move towards the outer surface making it positive. This positive charge will induce negative charge on the outer surface of the inner sphere and positive charge on the inner surface of the inner sphere. The ground connection removes these positive charges from the inner surface.
     
  14. Jan 23, 2010 #13
    Re: Capacitor

    Thanks a lot for your help !!
     
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