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7 November, 08:03

You need to design a capacitor capable of storing 3.0 10-7 c of charge. at your disposal, you have a 100 v power supply and two metal plates, each of area 0.490 m2 each. what is the limit of the separation of the plates? mm

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  1. 7 November, 10:34
    0
    Capacitance = Q/V

    Where Q is the charge

    V is the voltage

    C = 3 E - 7 C / 100V = 3 E - 9 F = 3nF

    Capacitance is also equal to

    C = (Eo*A) / d

    Where A is the area of the plate

    d is the separation between them

    Eo is the permittivity (8.85E-12 for Air)

    We find d

    d = (8.85E-12 * 0.49) / 3 E-9 = 1.4455 E - 3 m

    d = 1.4455 mm
  2. 7 November, 11:39
    0
    The capacitor formula is given by

    Capacitance = Q/V

    (3 * 10-7) / 100 = 3*10-9

    the relation between capacitance, Area and distance between the plates is given by

    C = (EoA) / d - - > (EoA) / C = d

    (8.85 * 10-12) * 0.490 / 3*10-9

    d=.1.44mm
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