# 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
140
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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
140
11
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
224
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I am curious...... Can you post the calculations...

4. ### Heavydoody Thread Starter Active Member

Jul 31, 2009
140
11
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.

5. ### Heavydoody Thread Starter Active Member

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

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