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Force between two spheres connected by wire and battery

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data
    There are two alike initially uncharged spheres of radius r at a distance of d>>r, as shown in the figure.
    4186fe470249e930eba0a6ece.gif
    a) Switch ##K_2## is closed. What is the force exerted between the two spheres?

    b) What would be the force between the two spheres, if in the previous experiment instead of closing switch ##K_2##, switch ##K_1## was closed?

    c) Determine the force exerted between the spheres if both switches are closed.

    2. Relevant equations


    3. The attempt at a solution
    In part a), if the battery sends a charge of ##Q## to the left sphere, then it must send a charge of -Q to the right sphere. Equating the potential difference gives:
    $$\frac{2kQ}{R}=U \Rightarrow Q=\frac{UR}{2k}$$
    Hence, the magnitude of force between the spheres is:
    $$F=\frac{kQ^2}{d^2}=\frac{kU^2R^2}{4k^2d^2}$$
    The force is attractive.

    Is this correct?

    In part b), I really have no idea. How do I deal with this case? :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Apr 16, 2014 #2

    rude man

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    part (a): right.

    part (b): can any charge ever accumulate on either sphere?

    part (c): what's the potential difference between the spheres now?
     
  4. Apr 16, 2014 #3
    I am not sure, it feels as if the charge won't ever reach to the right sphere given the connection is grounded in between but I think I am wrong.
     
  5. Apr 16, 2014 #4

    rude man

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    In order for charges to accumulate on the spheres, don't you need current to flow between them at some point, in this case displacement current? Can that happen when switch K2 is never closed?
     
  6. Apr 16, 2014 #5
    I am still very much confused. Also, I am not satisfied with what I did in part a). Even if I close the switch ##K_2## how the charge will ever reach the spheres (i.e a current should flow) when the circuit isn't even complete? :confused:
     
  7. Apr 16, 2014 #6

    rude man

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    The circuit IS complete. It's via the capacitance of the two spheres in mutual proximity.

    Are you at all electronics-oriented? You can answer all three questions easily by thinking of what happens in a capacitor connected by one lead by the + terminal of a battery and with the various connections defined in the three parts of your problem.
     
  8. Apr 16, 2014 #7
    Can I say that the two spheres represent two separate spherical capacitors? For both the spherical capacitors, the one at infinity is grounded so the circuit for part a) can be drawn as shown in attachment. Is it correct?

    I am not sure what do you mean by this.
     

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  9. Apr 16, 2014 #8

    rude man

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    No, the two spheres make up ONE capacitor. Each sphere is one "plate" of the capacitor.
     
  10. Apr 16, 2014 #9
    So in part b), there would be no force of attraction? For part b), only one plate is connected to capacitor so the circuit is open, no charge to the capacitor, right?

    What about c)? Can I redraw it as shown in the attachment?
     

    Attached Files:

  11. Apr 16, 2014 #10

    rude man

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    Right on all counts!
     
  12. Apr 16, 2014 #11
    Thanks rude man for all the help so far!

    Isn't part c) the same as part a) except the grounding connection? How would it affect the charges on the plates of capacitor? :confused:
     
  13. Apr 16, 2014 #12

    rude man

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    What's the potential between the spheres?
     
  14. Apr 16, 2014 #13
    The potential difference is U? Does that mean the force would be same as in part a)?
     
  15. Apr 16, 2014 #14

    rude man

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    Are you familiar with the First Uniqueness Theorem in electrostatics?
     
  16. Apr 16, 2014 #15
    No. :(
     
  17. Apr 16, 2014 #16

    rude man

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    Well, is the potential difference between the spheres any different than in part (a)?

    Hint: knowing V you can assume charge +Q and -Q on the spheres, use Gauss's theorem to determine the E fields generated by both spheres, integrate the force required on a unit test charge in going from the - sphere to the + sphere, equate to V, then that'll give you Q and of course then it's just the Coulomb calculation.
     
  18. Apr 16, 2014 #17
    No, its the same as in a).
    Before doing the math, isn't the condition in c) is same as a)? The spheres with charge Q and -Q are at potential kQ/R and -kQ/R respectively, so the potential difference, 2kQ/R, must be equal to U. This would give me the same answer as a). :confused:
     
  19. Apr 16, 2014 #18

    rude man

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    That's correct, but since you didn't know about the uniqueness theorem I thought you might want to confirm your conclusion a second way.
     
  20. Apr 16, 2014 #19
    Thanks rude man once again! :)

    I have seen the uniqueness theorem in books by Purcell and Griffiths but I used those books only for a short period of time so I couldn't gain much from them. Moreover, I think it involves a lot of math which isn't taught at a high school level.
     
  21. Apr 16, 2014 #20

    rude man

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    You're doing pretty impressive work for a high school student! Keep it up!
     
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