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Radius of variable capacitor

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A multi plate variable capacitor has 4 pair of plates. The plates, when closed, are separated in air by 0.01mm and a capacitance range of 10 to 400pF.

    a) Estimate the required radius R of each plate.

    b) The capacitor is set to maximum 400pF and is charged to 10V through a 50kΩ resistor. Determine:
    i) The initial value of the current flowing.
    ii) The time constant for the circuit.


    2. Relevant equations

    C = εr εo A (n-1)/d

    3. The attempt at a solution

    I am still new to all this so please dont go too technical with any replies.
    I think the correct formula to work out the radius is C=εr εo A (n-1)/d

    Transpose this to make A the subject:

    A= Cd/εr εo (n-1)

    Which gives me A= (400x10^-12)x(0.00001)/1 x 8.85x10^-12 x 7

    Area of plate = 0.00316 (I think this is meter^2)

    Then use this answer in A=∏r^2
    Transpose for radius = √A/∏

    Which gives me an answer of 0.0317 (Again I think this is meters).

    bi) The initial value of current flowing.
    If t=0 The charging current has its highest value.
    Therefore io=V/R
    =10/50000
    =0.0002 Amps

    bii) The time constant t=CR
    t=(400x10^-12)x 50000
    t=0.00002s

    Could you tell me if I am going about these the correct way or am I way off mark thanks.
     
  2. jcsd
  3. Dec 2, 2011 #2

    gneill

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    Staff: Mentor

    Presumably you mean (2n - 1) there, where n is the number of plates. In this case (2n - 1) = 7.
    I think you were tripped up by your own lack of parenthesis use. The "7" at the end of the calculation is part of the denominator for the last division.
    [tex] C = \frac{\epsilon_o A (2n - 1)}{d}[/tex]
    [tex] A = \frac{C\;d}{\epsilon_o (2n - 1)} [/tex]
     
  4. Dec 2, 2011 #3
    Thanks for that gneil
    Yeah the n is for the plates.
    Am I right in assuming the answers are in meters^2 for the first part?
    Could you also tell me if everything looks ok on the remaining answers.
    Many thanks
     
  5. Dec 2, 2011 #4

    gneill

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    The area will be in m2, yes. But the radius will be in meters.

    The remaining answers are correctly done. You might want to use scientific notation for the results, or express currents in mA and time in ms. It just looks neater!
     
  6. Jan 11, 2012 #5
    I get the radius to equal 6.41x10^-3. For question A, did you get this?

    I got area=6.457x10^-5

    Then half circle radius of plates = sqrt(2*6.457x10^-5)/(Pi)
     
  7. Jan 11, 2012 #6

    gneill

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    Your area value looks okay. Your formula for the radius does not. Can you show your derivation?
     
  8. Jan 11, 2012 #7
    Daft me..

    Should have been R=SQRT (6.457x10^-3)/(Pi)

    Answer 4.53x10^-3 m^2

    Apologies for my slack formulas.. Think this is correct now
     
  9. Jan 11, 2012 #8

    gneill

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    No, it still doesn't look right. Write out the (no numbers!) for the area given a radius r. Then show how you rearranged it to find r in terms of area.
     
  10. Jan 11, 2012 #9
    I think you mean;

    Area = Π*r^2
    radius = √(area/Π)
     
  11. Jan 11, 2012 #10

    gneill

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    Right. Note that the [itex]\pi[/itex] is inside the square root, which is not where you had it in your posts above.

    One more correction is necessary. The capacitor plates of the variable capacitor are in the form of half circles. When they completely overlap for maximum capacitance the total area of the opposing faces is that of a half circle.
     
  12. Jan 11, 2012 #11
    So I now get

    radius = √(6.457x10^-5/Π)
    =0.004533571 m

    However area = full circle so divide 6.457x10^-5 by 2 to get semicircle area

    Thus
    √(3.228x10^-5/Π)

    =0.00320

    Or 3.21x10^-3 m

    Sorry my physics is as poor as poor can be
     
  13. Jan 11, 2012 #12

    gneill

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    You're catching on, but it's still not quite right. The area of a plate must turn out to be equal to what you calculated earlier: 6.45 x 10-5 m2. After all, that's the area that the plates are required to have in order to achieve the required capacitance. Since the plate is in the shape of a semi-circle, the full circle would have double that area...
     
  14. Jan 11, 2012 #13
    I think I may be there...so..

    Double the area 6.457x10^-3
    To give 0.00012914

    radius =√(0.00012914/Π)

    Gives, 0.00641143 m or 6.41x10^-3 m

    This can be confirmed by;

    Area = Π*0.00641143^2 which gives the full circle area
    And divided by 2 gives 6.456x10^-5

    Hope this is it!!
     
  15. Jan 11, 2012 #14

    gneill

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    Yes, that looks better!
     
  16. Jan 11, 2012 #15
    Thanks for helping me, appreciate it
     
    Last edited: Jan 11, 2012
  17. Jan 11, 2012 #16

    gneill

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    Always happy to help! Good luck.
     
  18. Jan 11, 2012 #17
    Sorry to pop up again

    A second part asks;

    Having fully charged, the capacitor is then discharged through the 50 kΩ resistor. Determine:

    - the current flowing when the capacitor has been discharging for 5 μs

    I get ;

    I = (V/R)e^-0.25 = (10/5000)*(0.7788)=1.5576*10^-4=0.1557*10^-3=0.1557mA

    the voltage drop across the resistor when the capacitor has been discharging for 10 μs.

    I get;

    V=10exp(-0.5) = 10*(0.60653) = 6.0653volts

    Am I on the right lines with this? As youll be able to see I lack confidence in physics
     
  19. Jan 11, 2012 #18

    gneill

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    Yes, that's good.
     
  20. Jan 11, 2012 #19
    Thanks :)
     
  21. Jan 17, 2012 #20
    got it now, thanks
     
    Last edited: Jan 17, 2012
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