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Physics Experiment Help - Capacitors

  1. Oct 23, 2008 #1
    Note: This is a question about a physics experiment. As such, it does not completely fit into the provided template, but I'll do my best to stick to it anyway.

    1. The problem statement, all variables and given/known data
    The problem is about an experiment about parallel plates capacitors. The experimental setup is the following:

    1. Two round parallel plates with a radius = 0.0999 m.
    2. a constant current provided with magnitude 30uA
    3. The plates are charged and discharged at a circuit frequency of 20Khz using a solid-state switch
    4. -The resultant charging waveform is observed on an oscilloscope, and the potential difference between the bottom of the waveform and the top is recorded

    The full description of the lab can be found here: http://epic.mcmaster.ca/%7Eglen/2a3/2a4_exp3.pdf, if you're interested.

    I have several questions about this:
    firstly, there is a derivation in the lab manual of the capacitance with respect to the rate of change of voltage with respect to time:

    [tex]C = \frac{Q}{V}[/tex]
    [tex]Q = C[/tex]
    [tex]\Delta Q = C\Delta V[/tex]
    [tex]\frac{\Delta Q}{\Delta T} = C * \frac{\Delta V}{\Delta T}[/tex]

    but,

    [tex]\frac{\Delta Q}{\Delta T} = I [/tex]

    so we know that the V/T is inversely proportional to the capacitance.

    Question 1: re-arranging the equations leads me to believe this constant of proportionality is I, is that correct?

    I can use the following equation to calculate the theoretical capacitance of the plates:
    [tex] C = \frac{\epsilon A}{d} [/tex]
    Where A is the area of one of the plates (same as the other), [tex]\delta[/tex] is the permittivity of free space (~air), and d is the distance between the plates.

    So I calculated the theoretical capacitance & the capacitance from my measurements. I then proceeded to graph them:

    [​IMG]

    Obviously there's a bias error that is affecting my results.
    It seems to me that there are several things affecting these results, some of which I'm not sure what the exact effect of them is:
    1. The oscilliscope introduces resistive, capacitive and inductive loading
    2. the rest of the circuit has a small capacitance
    3. the frequency of the waveform (due to the current switching) leads to reactance (it will resist the change in voltage)?

    Question 2: It is unclear to me how exactly reactance & inductive loading tie into this DC current, I've only learned about reactance, impedance, etc in an AC context, could someone enlighten me?

    Question 3: Is my approach correct? Any further hints?

    Kind regards,

    Erik
     

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    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 23, 2008 #2

    Redbelly98

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    Yes.

    It may not be necessary to give a detailed analysis of all this in your lab report. Look at your graph, how (in simple terms) do the two curves differ?

    Your approach looks good. I suggest, however, that your graph's capacitance axis use either scientific notation or better yet unit prefixes (pF or nF). There are a lot of zeroes there that people must count to see just what the capacitance values are!
     
  4. Oct 24, 2008 #3
    Well, the only thing I can see is the bias error: the graph seems to be exactly the same, just shifted up by about 2nF. The circuit accounts for about 30pF of that (given). I figure the oscilloscope could add some to that (not quite sure how much).

    In addition, I'm curious what the effect of round plates is on fringing. 2*pi*r < 8*r, so I'd assume less fringing?

    Erik
     
    Last edited: Oct 24, 2008
  5. Oct 24, 2008 #4

    Redbelly98

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    I'm seeing a 0.2 nF or 200 pF shift between the graph's 2 curves, so you may want to double-check that number.

    The scope capacitance should be printed next to the input jacks, though I see you used a voltage probe and not a direct connection to the scope input jack. The probe would have a capacitance too.

    One thing I'm not sure of is how the stray inductance would affect the measurement.

    Not sure what you're comparing here. Circular vs. square plates? A good number to look at is the ratio of perimeters for the two shapes, assuming equal areas (same theoretical capacitance) for the circular and square plates.

    Good luck!

    Regards,

    Mark
     
  6. Oct 24, 2008 #5
    0.2 nF, yes. I made a mistake there.

    That's exactly what I was trying to compare. Your suggestion definitely makes alot more though. I'll look into it! Thanks for helping me out!

    Erik
     
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