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Homework Help: Amount of work that is done to adjust the capacitor from position A to position B

  1. Dec 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Flat capacitor with a plates distanced d = 2 cm and the area A= 1 dm² are connected to a voltage U = 100 V. We disconnect the source of voltage and we increase the distance between the plates to twice the distance. How much work is done?

    2. Relevant equations

    C= ε*A/ d
    Q= C*V
    E= ½ Q*V
    W= ΔE

    3. The attempt at a solution

    Calculating the capacitance at point A:
    d(A)= 0.02 m, A= 0.01 m², ε= 8.85 × 10^-12 As/Vm

    C(A)= ε*A/ d(A)
    C(A)= [(8.85 x 10^-12 As/Vm)*(1 x 10^-2 m^2)]/ (2 x 10^-2 m)
    C(A)= 4.43 x 10^-12 F

    Calculating the capacitance at point B:
    d(B)= 0.04 m, A= 0.01 m², ε= 8.85 × 10^-12 As/Vm

    C(B)= ε*A/ d(B)
    C(B)= [(8.85 x 10^-12 As/Vm)*(1 x 10^-2 m^2)]/ (4 x 10^-2 m)
    C(B)= 2.21 x 10^-12 F

    Calculating the charge on the plates:

    Q= C(A)*V(A)
    Q= (4.43 x10^-12)*100 V
    Q= 4.43 x 10^-10 C

    We know that the charge should be the same on both plate A and plate B so we use that piece of information to calculate the voltage of plate B:

    V(B)= Q/ C(B)
    V(B)= (4.43 x 10^-10 C)/ (2.21 x 10^-12 F)
    V(B)= 200 V

    Calculating the electrical energy in plate A:

    E(A)= ½ Q*V(A)
    E(A)= ½ (4.43 x10^10 C)* 100 V
    E(A)= 2.22 x10^-8 J

    Calculating the electrical energy in plate B:

    E(B)= ½ Q*V(B)
    E(B)= ½ (4.43 x10^10 C)* 200 V
    E(B)= 4.43 x10^-8 J

    Calculating the work:

    W= ΔE
    ΔE= E(B) – E(A)
    ΔE= 4.43 x10^-8 J - 2.22 x10^-8 J
    ΔE= 2.21 x 10^-8 J

    Are my calculations correct?:confused:
    Thank you for helping!:smile:
     
  2. jcsd
  3. Dec 29, 2009 #2
    Your calculations are correct. The distance of the plates compared to their size is too large
    to make the capacitance calculation so accurate however.
     
  4. Dec 29, 2009 #3
    :smile:Well, at least I cracked one problem!:biggrin:
    Thank you for your help and HAPPY NEW YEAR:wink:!
     
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