Anonymous
Anonymous asked in Science & MathematicsEngineering · 9 years ago

An air-filled parallel-plate capacitor has a capacitance of 1.6 pF.......?

An air-filled parallel-plate capacitor has a capacitance of 1.6 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.8 pF. Find the dielectric constant of the wax.

2 Answers

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  • 9 years ago
    Best Answer

    Please study this matter, Capacitance and Capacitors

    I. Elementary Characteristics

    In its most elementary state a capacitor

    consists of two metal plates separated by

    a certain distance d, in between the plates

    lies a dielectric material with dielectric

    constant έ = εoε , where εo is the dielectric of air.

    The dielectric material allows for charge to accumulate between the capacitor plates. Air (actually vacuum) has the lowest dielectric value of εo = 8.854 * 10-12 Farads/meter where the Farad is the unit for capacitance.

    All other materials have higher dielectric values, since they are higher in density and can therefore accumulate more charge.

    C = εoε * A/ d,

    In your question C1/C2 = ε1* d2/ε2*d1, or 1.6/2.8 = 8.854*10^-12 * 2/ ε2

    accordingly calculate the value of ε2

  • Jacob
    Lv 5
    9 years ago

    C = 1.6 pF = (epsilon*Area*k)/d

    rearranging,

    d = (epsilon*Area*k)/1.6pF

    where,

    k=1.0

    epsilon = 8.854*10^-12 F/m

    area = unknown

    doubling the separation,

    d = (2*area*1.0*8.854*10^-12)/(1.6*10^-12)

    d = 11.0675Area

    Now that d will be plugged into the new equation

    2.8 pF = (k*area*epsilon)/d

    2.8 pF = (k*area*epsilon)/(11.0675Area)

    areas cancel out, and you solve for k

    k = 3.5

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