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Stuck on the last problem.

  1. Feb 24, 2004 #1
    https://hw.utexas.edu/tmp/Muddam1/1077653690Xuj.pdf

    I did this: I manipulated [tex]E = \frac{q}{\epsilon*A}[/tex] to get [tex]q = E*\epsilon*A[/tex] (E= electric strength), so 1.6e7*8.85e-12*.0246*.0501 = 1.74516e-7 but because the HW asks the answer to the 1e-6, I divided 1.74516e-7/1e-6= .174516 but it's wrong does anyone know why?
     
    Last edited: Feb 24, 2004
  2. jcsd
  3. Feb 24, 2004 #2
    Your link does not work.
     
  4. Feb 24, 2004 #3
    Whoa....

    Ok this is the problem

    009 (part 1 of 2) 10 points
    A parallel-plate capacitor of dimensions
    2:46 cm £ 5:01 cm is separated by a 0:61 mm
    thickness of paper.
    Find the capacitance of this device. The
    dielectric constant · for paper is 3.7. Answer
    in units of pF.

    I got this right.

    010 (part 2 of 2) 10 points
    What is the maximum charge that can be
    placed on the capacitor? The electric strength
    of paper is 1:6 £ 107 V=m. Answer in units of 1e-6c.
    this is the one I need help on
     
  5. Feb 24, 2004 #4
    ...

    Never mind, I got it.
     
  6. Feb 24, 2004 #5
    So what's the answer?
     
  7. Feb 24, 2004 #6
    I used a different formula to get the answer: [tex]q = E*k\epsilon*A[/tex]
     
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