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Finding capacitance/dimensional analysis

  1. Mar 1, 2010 #1
    Two 2cm x 2cm square aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 100 V battery. What is the capacitance?



    Ok, C = Q/V = epsilon_o*A / (d)

    I used 0.02^2 for A (is that correct, or do I use the area of BOTH plates?), plugged in 0.00050m for d and the value for epsilon, and I got an answer that was on the order of 10^-12, and I know the answer is in picofarads, so that looked promising. However, when I did dimensional analysis, I couldn't make sense of the units I get from this calculation.

    So, am I crunching the numbers correctly, and if so (and in any case), can someone clarify how the units work out? Thanks!
     
  2. jcsd
  3. Mar 2, 2010 #2

    Matterwave

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    [tex][\epsilon]= \frac{As}{Vm}; [A]=m^2; [d]=m; \frac{[\epsilon][A]}{[d]}=\frac{Asm^2}{Vm^2}=\frac{As}{V}=\frac{C}{V}=F[/tex]

    All the variables inside square brackets [] denote the values, while all the variables outside the square brackets are the units (A=ampere, V=volt, m=meter, s=second, C=coulomb, F=farad)
     
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