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Capacitance of 3 Plates

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Three conducting plates, each of area A = 3.5 cm2, are connected in such a way that the center plate can be moved between the two fixed outer plates. If the distances between the outer plates and the center plate are d1 = 0.35 mm and d2 = 0.65 mm, find the effective capacitance C of the system.

    2. Relevant equations
    C = E0*A/d

    3. The attempt at a solution
    I have tried adding the Capacitance of each plate (using equation above) with d1 and then d2 and finally d3 (total distance). I added that together but can't seem to get the right answer. I think it confuses me that there are 3 parallel plates because all the examples we talked about in class have been with 2 parallel plates.
     
  2. jcsd
  3. Apr 11, 2015 #2

    gneill

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    Staff: Mentor

    Hi Angie K., Welcome to Physics Forums!

    The plate in the middle provides one plate for two separate capacitors. Effectively then, you have two capacitors in series.
     
  4. Apr 11, 2015 #3
    So I would add the 2 plates on the sides (in series) to the plate in the middle which is in parallel to the outer two plates?
     
  5. Apr 11, 2015 #4

    gneill

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    It takes two plates to make a parallel plate capacitor. Each outside plate forms a capacitor with the plate in the center:

    Fig1.gif
     
  6. Apr 11, 2015 #5
    So my equation to solve for the effective capacitance of the system should look like this:

    C = E0*A/d1 + E0*A/d2
    where E0 is (8.85*10^-12)
    d1 is the given distance of .35 mm (convert to meters)
    d2 is the second given distance of .65 mm (convert to meters)

    Am I understanding this correctly?
     
  7. Apr 11, 2015 #6

    gneill

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    Not quite. How do capacitors in series "add"?
     
  8. Apr 11, 2015 #7
    To add the in series, I would add the reciprocal of the Capacitors 1/Ctotal= 1/C1+1/C2+1/C3... etc
     
  9. Apr 11, 2015 #8

    gneill

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    Yup.
     
  10. Apr 11, 2015 #9
    I really appreciate your help, I was just not understanding the problem correctly. Thank you for your clarifications.
     
  11. Apr 11, 2015 #10

    gneill

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    You're welcome :smile:
     
  12. Apr 11, 2015 #11
    I am still not getting the right answer. Since the plates are in series, here is what I did:

    (1/C=1/E0*A/d1) + (1/C=1/E0*A/d2)

    ? Where did I go wrong in my calculations
     
  13. Apr 11, 2015 #12

    gneill

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    Can you show some details of your calculations? What values are you getting for the individual capacitances? How about the net value?
     
  14. Apr 11, 2015 #13
    1/C1 = 1/(8.85*10^-12)*(.035m)^2/(.00035m) = 3.2284*10^10
    1/C2 = 1/(8.85*10^-12)*(.035m)^2/(.00065m) = 5.9956*10^10
    1/C1+1/C2 = 9.2240*10^9
     
  15. Apr 11, 2015 #14

    gneill

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    You need to check your plate area value. The conversion to square meters doesn't have the right order of magnitude.
     
  16. Apr 11, 2015 #15
    A = 3.5 cm^2 is .00035m^2 which is 1.225*10^-7 right?
     
  17. Apr 11, 2015 #16

    gneill

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    .00035m^2 is correct. But I don't understand where the 1.225*10^-7 comes from or what it's supposed to represent o_O
     
  18. Apr 11, 2015 #17
    Well, when I put .00035m^2 (.00035^2) in my calculator, it came out as 1.225*10^-7
     
  19. Apr 11, 2015 #18

    gneill

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    Ah. .00035m^2 is already the area in square meters (The units m^2 tell you so). You don't want to square it again!
     
  20. Apr 11, 2015 #19
    So then:

    1/C1 = 1/(8.85*10^-12)*(.00035m)/(.00035m) = 1.129E11
    1/C2 = 1/(8.85*10^-12)*(.00035m)/(.00065m) = 2.0984E11

    1/C1+1/C2 = 3.0984E-12
     
  21. Apr 11, 2015 #20

    gneill

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    Okay, those numeric values look good.
    While the numerical value is correct for the net capacitance, the equation that you wrote is not correct! 1/C1 + 1/C2, given the above values, is 3.23E11 F-1. So you've not shown that you took the reciprocal. When you show mathematical expressions, be sure that what you present is precise.

    Also, be sure to include units on numerical results. If you don't, a marker will deduct points or simply declare the answer to be incorrect.

    So summarizing, you've arrived at a correct numerical value but need to clean up the presentation and include units on your result.
     
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