Capacitance of a non-parallel plates

Discussion in 'Physics' started by Manu Bajpai, Feb 26, 2016.

  1. Manu Bajpai

    Thread Starter New Member

    Feb 25, 2016
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    I'm looking for the analytical approach for quite some time but to no avail. The angle between the plates may be theta and the plates may have area A.
     
  2. #12

    Expert

    Nov 30, 2010
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    Try this: Imagine one plate horizontal and you are looking at the edge of it. Another plate is above the first plate. If it is horizontal the answer is the obvious one. As the upper plate tilts upwards at the left edge, the area of the second plate exposed to the first plate changes as the cosine of the angle measured from the stationary edge to the horizontal plane. At the same time, the distance of the upper plate changes so the nearest edge is at the original distance but the farther edge is farther away. You have both diminished area and increased distance.
     
  3. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I am not entirely sure if this is the same idea as #12 had in the previous post but i will be stating this differently anyway so let's see what happens :)

    First, i assume you know the capacitance of a truly parallel plate capacitor.
    Second, rather than find a new formula from somewhere let's see if we can develop one right from the definition of a parallel plate capacitor. Well, let's see if you can do it, if you want to that is. Note that there are other ways but this one i think is more interesting.

    To start, we have the capacitance of a parallel plate capacitor in simplified form:
    C=K*A/d

    where
    A is area,
    d is distance between plates,
    K is a constant which we assume does not change with distance.

    We will adhere to the above in order to keep this simple.

    To start this off we divide the plate area in half, say across the width, and keep each plate parallel, and the distance for each plate will be the different for each of these two plates. That gives us two parallel plate capacitors connected in parallel. The distance between the first two plates d1 is the distance of the left edge of the original cap, and the distance of the second two plates d2 is the distance of the right edge of the original cap. So the combined capacitance of this new structure is:
    Cp=(A/2)*K/d1+(A/2)*K/d2

    Simplifying this a little we get:
    Cp=((d2+d1)*A*K)/(2*d1*d2)

    That is a rough approximation of what the final value of the cap will be. It's very rough because we only used two caps to emulate the original slanted plate capacitor.

    We can use more partial segments, so if we use 3 we get:
    Cp=((d2*d3+d1*d3+d1*d2)*A*K)/(3*d1*d2*d3)

    and that is a better approximation than we had with just two plates.

    If we use 4 we get:
    Cp=((d2*d3*d4+d1*d3*d4+d1*d2*d4+d1*d2*d3)*A*K)/(4*d1*d2*d3*d4)

    and that is even better than with just 3 plates.

    So you can start to see how this helps to visualize what is happening with the slanted plate, it is like a variable capacitor where the capacitance varies across the width, and each one is in parallel with the next so the capacitance adds. The result with an infinite number of divisions (infinite number of parallel plate caps) would equal the true result of the capacitance of the slanted plate cap.

    Intuitively, this is fairly easy to imagine. We are simply cutting the cap into little pieces and computing each part with a formula we already know. When we cut it into small enough pieces, we get a reasonably accurate result.

    Mathematically however, this is not in the simplest form. We would have to convert this into a more convenient math form and then integrate over all pieces. That would lead to an exact value also.

    I would not be able to live with myself if i did not repeat the fact that very few calculations involving real capacitors will be truly exact, and that is because the electric field does not conform to what we would like it to be which would be a field that is completely confined between the two plates, even if they are really parallel. In fact, the field tends to spread at the edges of the cap and thus that changes the overall calculation. We usually neglect this however in problems like this for simplicity, because we usually dont want to jump into deeper physics just to get the main idea. That, and even more advanced physics wont be exact when we consider the construction of the cap will not be perfect either even given a well equipped machine shop.

    A numerical example using unity values for several constants (K, width, length) and distances 1 and 2 at the ends of the slanted plate i get the following results:

    One slanted plate, exact calculation: 0.6931
    One parallel plate, using average distance: 0.6667
    Two parallel plates, using two average distances: 0.6857

    The one parallel plate averaged approximation comes within about 4 percent of the exact value.
    The two parallel plate averaged approximation already comes within about 1 percent of the exact value.
    Not bad for a relatively simple calculation, and would be good for finding out if a more advanced calculation came within the ball park of the right result.

    Units above are in scaled Farads (scaled because of the unity constants used for simplicity). The scale factor involves the static electric space constant and the dielectric constant.
     
    Last edited: Feb 26, 2016
    anhnha likes this.
  4. #12

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    Nov 30, 2010
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    That's basically how Calculus works, but my education is deficient so I tried to say it in language no higher than trigonometry.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi there,

    You are right and if we did the calculus version it would be sort of the same, and we'd end up with the formula for a capacitor with one slanted plate that contains one or two logarithms. The simpler calculus version converts everything into functions of x, where x runs along the width (or length). It's pretty cool how it works out.

    But what did surprise me was that using the average plate distance (which would be the distance from the very center of the slanted plate to the center of the flat plate) came out within 4 percent of the exact calculation. That was using somewhat extreme values for the slant too, where one side was 2 times higher than the other. Maybe other ratios wont turn out so good though :)

    What else is a little interesting is that the slanted plate has to be longer than the flat plate in order to reach across the span of the flat plate while it makes the angle.
    If we slanted both plates then we can keep them the same size if the angle is the same for both plates.
     
  6. #12

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    Nov 30, 2010
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    Instinctively, I think the capacitance is related to 1/tan theta.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    This might work, but we are neglecting fringing fields and those might become dominant as we progress along this train of thought. This approach is also assuming that the electric fields in the dielectric are always a straight line, which we know is not the case since they will leave each plate normal to that plate.
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Yes, we'd have to include the angle too if we really wanted to relate the angle to the overall capacitance. I just kept it short and let the OP do some of the work.
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    Yes that's certainly true and i agree fully, that's why i included this note 5th paragraph from the end in post #3"

    "I would not be able to live with myself if i did not repeat the fact that very few calculations involving real capacitors will be truly exact, and that is because the electric field does not conform to what we would like it to be which would be a field that is completely confined between the two plates, even if they are really parallel."

    I also even mentioned that the physical construction itself can not be made to be perfect, and even a true parallel plate capacitor wont have perfectly formed plates (warped, pitted, etc.) so even if we used a more advanced calculation (field shape non uniform) we still cant get to be truly exact because we would then be assuming a plate shape of a certain type.
    So i was trying to make it clear that even a parallel plate capacitor will have a field that is not exactly how we normally assume it will be for these somewhat simplified problems. These approximations are often acceptable for a lot of course work and when not appropriate the problem statement usually mentions that, and you'll notice that a lot of solutions we find on the web come out to exactly the same formula we find on the web in different places. I am guessing that maybe 4th or 5th year college they would want more detail.
    We could go through the calculation but i am betting that this student does not need that level of perfection. Interesting to think about though, as we imagine the 'lines' going from one plate to the other. I wonder what happens if a mosquito flies in between the plates :)
     
  10. #12

    Expert

    Nov 30, 2010
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    It's called a, "bug zapper".:p
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello again,

    Ha ha, yeah that makes sense :)
    At a place i worked one time long time ago, we built one using an old CRT flyback transformer, and a grid of wires, mainly for catching flies. If the fly landed wrong it would only get hurt not killed. I think the voltage needed to be a bit higher.

    Back on topic...
    I also should mention that if we use the fact that the 'ideal' lines would look like circular arcs, that should only change the apparent distance (the electrostatic distance rather than the Euclidean distance) and that would look like a constant making all the distances (across the width say) look larger. So the capacitance would be less than the linear method. However, i dont think that yields the best result either as the last i have read says that the only trusted method for a real life construction is the FEM (Finite Element Method). I havent yet used FEM for an electrostatic problem so i'd have to look into that in order to find out how to proceed. The arc method should be easy though as each distance is simply changed according to the length of the arc rather than the vertical distance from plate to plate. That might be good enough for very thin plates though (thicker plates have edges that can have a significant effect). Perhaps i'll generate a formula unless anyone else cares to do it.
    The linear formula looks like this:
    C=K*W*L*(ln(d2/d1)/(d2-d1))
    with d2 the greater plate separation, d1 the lesser, and W the width and L the length. K is the combined permittivity (free space and relative). This comes from finding an expression for dC/dw and then integrating over the width w.
    Since d1 and d2 are related by trigonometry, it should not be too hard to get that into a form with angle 'a' rather than two distances.
     
    Last edited: Feb 28, 2016
  12. Glenn Holland

    Member

    Dec 26, 2014
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    Smells like an integration problem.

    However with large distances and angles, the path of electric and magnetic fields can be quite erratic. In your case, there's so much fringing that a finite element analysis would be the only way to get an accurate result.
     
  13. Lool

    Member

    May 8, 2013
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    I think you will find this paper useful.
     
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