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Capacitor with two angled dielectric materials

  • Thread starter Siberion
  • Start date
  • #1
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Homework Statement


Find the capacitance of the capacitor shown in the figure. Asume H << L.
http://imageshack.us/a/img39/1935/capacitor.png [Broken]

Homework Equations



Capacitance of a plate parallel capacitor with a dielectric material e: eA/H where A denotes the area of the plate, and H the separation between plates


The Attempt at a Solution


My thoughts:

The line separating the dielectric materials can be written as f(x)=Hx/L (I rotated the system so the line starts in the origin)
The system can be considered as multiple parallel capacitors in which the proportion of the dielectric changes at the rate given by the linear equation.
First, I calculated the differential capacitance of two capacitors in series, whose lenght is dx. Their capacitances are given by:

[tex]C1= \frac{e1Ldx}{H-\frac{Hx}{L}}[/tex]
[tex]C2= \frac{e2Ldx}{\frac{Hx}{L}}[/tex]

After getting the equivalent capacitance, I integrated from 0 to L, hoping to get the total capacitance of the system.
Thing is, when I assume e1=e2, I don't get the capacitance of a normal plate parallel capacitor with one dielectric material.

I really want to know if the logic of my process is okay, because I've tried about 5 times to check if calculus are done right.

Sorry for my english and the simplicity of my explanation, I'll try to explain further when I get a little more time.

Thanks in advance!
 
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Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Your method is correct. Checking the final answer for e1 = e2 could be a little tricky because it might require taking a limit as e1 approaches e2. An easy thing you can do is let e1 = e2 before doing the integral and see if the integral gives the expected result. If not, you know you haven't set up the integral correctly.

It would help if you would show the form of the integral that you set up. From your expressions, it appears that the capacitor plates are squares of edge length L.
 

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