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Multi-Plate variable Capacitor Questions.

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A multi-plate variable capacitor has 4 pairs of plates. The plates, when closed, are seperated in air by 0.01mm. A capacitance range of 10 - 400 pF is required

    a) estimate the required radius, R of each plate.

    b) The capacitor is set to the maximum of 400 pF and is charged to 10v throught a 50KΩ resistor. determine: i) the initial value of current flowing. ii) the time constant for the circuit.


    IMAG0096.jpg


    2. Relevant equations

    capacitance = εr ε0A(n-1) / d

    Initial current = V/R

    Time constant = R x C

    3. The attempt at a solution

    a) 400x10-12 = εr ε0A(8-1) / (1x10-5)

    can anyone explain what these 2 values are? (εr & ε0) i think one is dialectric strength. which for air is 3MV m-1 ?

    b)
    i) 10V/50Kohm = 0.2mA

    ii) (50x103) x (400x10-12) = 20 μs
     
  2. jcsd
  3. Mar 27, 2012 #2

    gneill

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    Staff: Mentor

    ##\epsilon_o = 8.8542\times10^{-12}\;F m^{-1}## is the permittivity of vacuum (free space)

    ##\epsilon_r = 1.00058986## is the relative permittivity for air
     
  4. Mar 27, 2012 #3
    Ok thanks, so...


    3. The attempt at a solution

    a) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


    re-arranging for A.

    A = C.d / ϵor


    A = (400x10-12)*(1x10-5) / (8.8542x10-12)

    A = 0.451763x10-3 = 0.451763 mm2


    A= ∏r2

    r = √(a/∏)

    r = √(0.451763x10-3/∏) = 11.99mm

    can anyone check if this is ok?


    b)
    i) 10V/50Kohm = 0.2mA

    ii) (50x103) x (400x10-12) = 20 μs


    do these answers look plausible?
     
    Last edited: Mar 27, 2012
  5. Mar 27, 2012 #4

    gneill

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    Staff: Mentor

    Oops. What happened to the "(8 - 1)" ?
    Are the plate areas full circles?
    These last two look fine.
     
  6. Mar 27, 2012 #5
    oops forgot about that. got carried away! :D


    a) 400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


    re-arranging for A.

    A = C.d / ϵor(n-1)


    A = (400x10-12)*(1x10-5) / (8.8542x10-12)(7)

    A =0.0645376x10-3 = 0.06454mm

    A= ∏r2

    r = √(a/∏)

    r = √(0.0645376x10-3/∏) = 4.532mm


    they are half circles

    so if i multiply the Area by 2.

    0.064545x10-3*2 =0.1290752x10-3

    then work out the radius:

    r = √(0.1290752x10-3/∏) = 6.4098mm
     
    Last edited: Mar 27, 2012
  7. Mar 27, 2012 #6

    gneill

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    Staff: Mentor

    I think you have a unit conversion problem; The area should be three orders of magnitude larger if it's in mm2 (and it's mm2, not mm as you've written). Note that an area of 0.06 mm2 would be practically microscopic! That should be a tipoff that something's amiss :wink:
    Again, are the plates full circles?
     
  8. Mar 27, 2012 #7
    ok. i'm not really sure where i'm going wrong here.



    a)

    C = (ϵor(n-1)) / d

    C = 400x10-12 Farads
    ϵo = 8.8542×10−12Fm−1
    ϵr = 1.00058986
    d = 1x10-5 m


    400x10-12 = 1 x (8.8542x10-12)A(8-1) / (1x10-5)


    re-arranging for A.

    A = C.d / ϵor(n-1) is equation ok?



    A = (400x10-12)*(1x10-5) / 7(8.8542x10-12)

    A =0.066896x10-3 mm2
     
  9. Mar 27, 2012 #8

    gneill

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    Staff: Mentor


    Yes, that's fine.
    A = 6.45 x 10-5 m2

    6.45 x 10-5 m2 x (106 mm2/m2) = 64.5 mm2
     
  10. Mar 28, 2012 #9
    ok thanks for that. I've never been great with unit conversions. :blushing:



    so now do I multiply the calulated area by 2 because the plates are semicircles?

    A = 64.5376mm2*2 = 129mm2


    then work out the radius....

    r = √(129/∏) = 6.41mm
     
  11. Mar 28, 2012 #10

    gneill

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    Staff: Mentor

    That works fine. You could also have said that each plate is a semicircle so it has area

    ##A = \frac{1}{2}\pi r^2##

    It works out to the same thing.
    Looks like you're done :smile:
     
  12. Mar 28, 2012 #11
    many thanks for your help and patients.
     
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