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Question based on capacitor

  1. Jul 24, 2013 #1
    1. The problem statement, all variables and given/known data
    See the image:
    http://postimg.org/image/q550yqqd1/ [Broken]

    Left click to enlarge....

    2. Relevant equations

    For a parallel plate capacitor, Capacitance C=KεoA/d

    A is area of plate and d is distance between the two plates..

    3. The attempt at a solution

    I do not know how to approach. I divided the capacitor along the diagonal into two halves. For the first half,

    dC1 = K1εodA/dl

    I took a strip of edge dx and length "l" perpendicular to the capacitor. Area of that strip= (x+dx)2-x2=2xdx..

    dA=2xdx

    Hence,
    dC1 = 2K1εoxdx/dl

    Now taking an angle θ, tanθ=d/a= (d-l)/(a-x)....

    Am I going in the right direction ?

    Please help !!

    Thanks in advance...
    :smile:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 24, 2013 #2

    gneill

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    Staff: Mentor

    I can't clearly picture how you're dividing up the device. In my mind's eye I can see taking vertical slices of width dx across the image. At some horizontal position x, a slice will effectively consist of two capacitors in series, one with dielectric K1, the other with dielectric K2. "Plate" area is given by a*dx, and "plate" separation for each capacitor can be determined with a bit of geometry.
     
  4. Jul 25, 2013 #3
    Ok, So I have a differential form,

    dC2=K2εoadx/distance, as per the hint you gave. I also understand that the two capacitors are in series. So what should I replace K with ?

    And about geometry.... I took angle θ between diagonal and an edge. Then,

    tanθ = d/a = y/x = (d-y)/(a-x)

    where, the capacitor with dielectric constant K2 is positioned at distance x and its distance from the one edge(horizontal) to the boundary of second capacitor.

    Am I on the right track ? Thanks..
     
  5. Jul 25, 2013 #4

    gneill

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    Staff: Mentor

    I believe so... can you write expressions for the two capacitances of a slice?
     

    Attached Files:

  6. Jul 26, 2013 #5
    gneill,

    As per the image you gave,

    dC1 = K1εoadx/d1

    dC2= K2εoadx/d2

    Correct ?

    Now dC=dC1*dC2/(dC2+dC1) ?
     
  7. Jul 26, 2013 #6

    gneill

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    Staff: Mentor

    Right. You'll want to express d1 and d2 in terms of some common variable over which you can integrate.
     
  8. Jul 27, 2013 #7
    Ok, so replacing d2=d-d1,

    dC1 = K1εoadx/d1

    dC2= K2εoadx/(d-d1)

    Now, dC=dC1*dC2/(dC2+dC1)
    dC=K1K2εoadx/{K1(d-d1)+K2d1}

    Now to what limit shall I integrate this expression over ? Looks like I will have to eliminate d1.

    By geometry, I got,

    d(a-x)/a = d1

    Correct ?

    P.S. d is distance between plates as given. dC is small capacitance. dx is small distance between slices as per your image. (Thanks.)
     
    Last edited: Jul 27, 2013
  9. Jul 27, 2013 #8

    gneill

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    Staff: Mentor

    Sure. So you're integrating w.r.t. x. You should be able to see from the diagram the limits for x.
     
  10. Jul 27, 2013 #9
    Thanks a lot gneill !!!! :smile:

    I integrated the expression in limits of x from zero to "a" and got the correct answer !! Thanks once again.
     
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