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Concentric Spherical Capacitors

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Figure P26.26 shows six concentric conducting spheres, A, B, C, D, E, and F having radii R, 2R, 3R, 4R, 5R, and 6R, respectively. Spheres B and C are connected by a conducting wire, as are spheres D and E. Determine the equivalent capacitance of this system.

    2. Relevant equations
    Capacitance of a sphere: C = 4pi(epsilon0)R
    Capacitance in general: C = Q/V


    3. The attempt at a solution
    I’m supposed to find the equivalent capacitance of this arrangement. I was kind of perplexed by this one. But here’s what I’ve been thinking. OK. B and C are at the same potential (because they're connected by a wire). D and E are also at the same potential. So it’s like B and C are in parallel, as are D and E. So I said CBC = CB + CC, and CDE = CD + CE. And then CEQ = 1/(1/CA + 1/CBC + 1/CDE + 1/CF) (since I considered these to be in series). Capacitance of a sphere is C = 4πεOR. So using this approach, I found CEQ = (90/133)( 4πεOR). Do I have the right idea, or am I way off?
     

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  3. Jul 13, 2009 #2
    I don't think your equation for spherical capacitance is correct. See for example http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capsph.html

    Additionally, I'm pretty sure that the equation you used for equivalent capacitance only applies to parallel-plate capacitors which are separated.

    I think the tricky thing here would be using the correct radius in your calculations.
     
  4. Jul 13, 2009 #3

    turin

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    Homework Helper

    The question is ambiguous. Capacitance is a relative property between two conductors. Between which two spheres are you to determine the capacitance?

    Alternatively, the capacitance can refer to a single conductor relative to zero charge at an infinite distance (which is the origin of the formula C = 4πεR). In this alternative interpretation of capacitance, you still need to refer to a particular sphere.

    All six spheres taken together have an undefined capacitance (becuase there are more than two disconnected conductors).
     
  5. Jul 13, 2009 #4
    OK, that may very well be the case. I just feel like there must be a solution since it is one of the problems in Physics for Scientists and Engineers with Modern Physics (5th Edition) by Raymond A. Serway and Robert J. Beichner (Chapter 26, Problem 26). So I guess I'm stuck. Maybe I'll just skip this problem--I'm not doing this for a class, I'm just working on compiling a complete set of solutions for the homework problems in this 2-volume set. Thanks for your help.
     
  6. Jul 13, 2009 #5
    I'm going to take a guess at this one, but it looks to me that you have three spherical capacitors connected in series. Capacitor 1 has radii a&b, connected in series to capacitor 2 with radii c&d which is connected in series to capacitor 3 with radii e&f. The capacitance is dependent on the radius C=a*b/(Ke*(b-a)) where a is the inner radius and b is the outer radius. I would like to see the book answer if there is one.
     
  7. Jul 13, 2009 #6
    That's an interesting idea. I see what you mean with the three capacitors connected in series. Actually, that seems like a great idea! So let's say I put +Q on sphere A. Then sphere B must have -Q so that Gauss's Law gives 0 E-field between spheres B and C (otherwise charges would flow in the wire connecting B and C). However, since the combination of B and C was initially uncharged, then C must now have +Q. This means that D has -Q (so again Gauss's Law gives an E-field of 0 between D and E so no current flows in the wire). And thus E has +Q, and finally F will have -Q. However, now that it seems, as you said, that we have 3 capacitors in series, how do they add? I guess charge is the same on all 3 capacitors (since they are in series), so is it true that 1/Ceq = 1/Cab + 1/Ccd + 1/Cef? I think you might be on to something! Thanks a lot!
     
    Last edited: Jul 14, 2009
  8. Jul 13, 2009 #7
    P.S. Using this approach, I found Ceq = 4πεO(60/37)R.
     
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