Capacitance for Nonparallel Plates

Discussion in 'Homework Help' started by Heavydoody, Feb 21, 2010.

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  1. Heavydoody

    Thread Starter Active Member

    Jul 31, 2009
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    This is an extra credit homework problem for my physics class. Two square plates, 5cm on a side, are set at an angle of thirty degrees. The leading edges of the plates begin at a radius of 1cm from the center of the angle. What is the capacitance?

    I believe I will have to integrate along the length such that dx becomes the width of plates for individual little capacitors. Thus, I have the integration of (.05dx/(2xsin15)) from .01 to .06 times ε. Does this sound like the right approach? I get approximately 153pF.
     
  2. Heavydoody

    Thread Starter Active Member

    Jul 31, 2009
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    Well, for anyone interested...when I finished my calculations, I actually came up with 1.53pF...we turned it in today and the prof gave us the solution...turns out I was close. I needed to take the arc length instead of the straight line distance. The answer ended up being 1.51pF. I am happy that I was able to apply integration in a practical situation though, as I am still not completely comfortable with calculus.
     
  3. lmartinez

    Active Member

    Mar 8, 2009
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    I am curious...... Can you post the calculations...:rolleyes:
     
  4. Heavydoody

    Thread Starter Active Member

    Jul 31, 2009
    140
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    This is what it looked like. The way it was explained was that not only were we integrating along the radius (what I was calling x), but we also had to take θ down to where the tiny plates were approximately parallel, hence the arc length as opposed to the straight line.

    [​IMG]
     
  5. Heavydoody

    Thread Starter Active Member

    Jul 31, 2009
    140
    11
    ...and my (incorrect) solution:

    [​IMG]
     
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