Anonymous
Anonymous asked in Science & MathematicsEngineering · 9 years ago

A parallel-plate air-filled capacitor has a capacitance of 43 pF.?

A parallel-plate air-filled capacitor has a capacitance of 43 pF. (a) If each of its plates has an area of 0.28 m2, what is the separation? (b) If the region between the plates is now filled with material having κ = 8.2, what is the capacitance?

3 Answers

Relevance
  • Jacob
    Lv 5
    9 years ago
    Favorite Answer

    43 pF = 0.000043 microF

    using the equation C=(k*epilison*Area)/separation

    where,

    epsilon = 8.854*10^-12 F/m

    C= 43 pF

    A=0.28

    k=1.0

    solve for d (separation)

    d=5.765 cm

    b)

    C = (8.2*0.28*8.854*10^-12)/(0.05765)

    C = 3.525*10^-10 F = 352.5 pF

  • 4 years ago

    This Site Might Help You.

    RE:

    A parallel-plate air-filled capacitor has a capacitance of 43 pF.?

    A parallel-plate air-filled capacitor has a capacitance of 43 pF. (a) If each of its plates has an area of 0.28 m2, what is the separation? (b) If the region between the plates is now filled with material having κ = 8.2, what is the capacitance?

    Source(s): parallel plate air filled capacitor capacitance 43 pf: https://shortly.im/7KE7V
  • 4 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

Still have questions? Get your answers by asking now.