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Field strength - capacitor.

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data

    The insulation in a plate capacitor consists of two layers with the thickness of 1 and 2 mm, and a relative permittivity of 4 and 6. Calculate the field strength in the thinnest layer when the capacitor is connected to a voltage of 5000v.

    2. Relevant equations



    3. The attempt at a solution

    I know that the answer should be in voltage per meter, and I find the following equation to be relevant:

    E = [itex]\frac{F}{Q}[/itex] = [itex]\frac{k*Q}{r^2}[/itex]
    where k is a constant.

    My problem is the numbers I'm given. Charge density divided by permittivity equals field strength? I don't know what charge density even is (a formula I found while google'ing the problem).

    Can anyone guide me as to where I should begin solving this problem?

    Appreciate any input.

    - Mutaja.
     
  2. jcsd
  3. Mar 2, 2014 #2

    gneill

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    Staff: Mentor

    A capacitor with multiple layers of dielectrics can be modeled as a series of capacitors, one for each layer. As a suggestion, start with the formula for the capacitance of a parallel plate capacitor and think "voltage divider".
     
  4. Mar 2, 2014 #3
    Formula: E = [itex]\frac{σ}{ε}[/itex] where σ = charge density and ε = permittivity. As stated in my first post.

    σ = [itex]\frac{Q}{A}[/itex] where Q = charge on plate and A = plate area.

    This whole concept of capacitors is very new to me, at least this aspect of it.

    I also know that the voltage difference between the two plates can be expressed in terms of the work done on a positive test charge q when it moves from the positive to the negative plate.

    V = [itex]\frac{work done}{charge}[/itex] = [itex]\frac{Fd}{q}[/itex] = Ed.

    I don't know the area of the plates, I don't know the distance between them. All I know is their thickness, relative permittivity (dielectric?) and the voltage used. Can I somehow use their relative permittivity to work out, for example, the distance or area?
     
  5. Mar 2, 2014 #4

    gneill

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    The distance between plates is set by the thickness of the dielectrics. Assume the same plate area for all, and represent it by a variable. It'll cancel out, so no worries.

    You must have seen the formula for the capacitance of a parallel plate capacitor:
    $$C = \epsilon \epsilon_o \frac{A}{d}$$
    where ##\epsilon## is the relative permittivity, A the plate area, and d the plate separation. You're given the thicknesses of each capacitor. So write expressions for the capacitance of each capacitor leaving A as an unknown.

    attachment.php?attachmentid=67171&stc=1&d=1393773577.gif

    If you consider the equivalent circuit (far right of above figure) you should be able to see that there's a capacitive voltage divider. Can you write an expression for the voltage across the thinner capacitor (C1)? Then you'll have the simpler case of one capacitor with dielectric and potential difference to deal with.
     

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  6. Mar 2, 2014 #5
    I must've missed something, somewhere. Yes, I have the equation $$C = \epsilon \epsilon_o \frac{A}{d}$$ in my book. But both C and A is unknown? Also, you're saying that A is cancelled out first, because it's the same area for both capacitors, but then I'm supposed to to rewrite the equation leaving A as the known? I still have no clue what C is because I need the A. Also, what is d?!

    I thought dielectrics was the same as permittivity somehow. I'm at a total loss here. I will try to clear my head and give it a go again soon, because right now I'm clueless.

    Sorry.
     
  7. Mar 2, 2014 #6

    gneill

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    A is an unknown that will eventually disappear once you've written the voltage divider expression. C is the capacitance of the individual capacitors and is what you want to write the expressions for using your equation above. d is the distance between plates. The d's for each capacitor are given to you in the problem statement as the thicknesses of the dielectric layers.

    A dielectric is a material with certain electrical properties, the important one being its permittivity.
     
  8. Mar 2, 2014 #7
    Ok, I now get WHY A is cancelled out. What I don't understand is how. I've tried setting up an expression for voltage dividing, and I realize why that's a good option - but I can't see how to do it.

    For my expression for capitance for each of the capacitors, how does the relative permittivity "work"? I know permittivity as a constant. 1 for vaccume, 4 for oil etc. Relative means compared to something else in my head.
     
  9. Mar 2, 2014 #8

    gneill

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    Suppose that the thinner capacitor is called C1 and the other C2. Write the voltage divider expression for the voltage across C1.

    Then substitute your capacitance expressions for C1 and C2. Simplify.

    Yes, it's a unitless scaling factor for the permittivity. It tells you how much larger (or smaller) the material's permittivity is compared to that of vacuum (##\epsilon_o##). It should be clear from its location in the parallel plate capacitance formula.
     
  10. Mar 2, 2014 #9
    Am I onto something here?

    VC1 = [itex]\frac{4*\frac{A}{d}}{4*\frac{A}{d}+6*\frac{A}{d}}[/itex]

    Since ε0 = 1.

    IF this is even remotely correct, I'm still confused about the d.

    I'm really trying here, but honestly, this is the best I can do. Thank you so much for staying with me trying to figuring this out.
     
  11. Mar 2, 2014 #10

    gneill

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    Each capacitor has its own d. It's the thickness of its dielectric. That should be apparent from the diagram I posted above.

    ε0 is not 1. It's the permittivity of free space (vacuum) with units of Farads per meter, so
    $$\epsilon_o = 8.854 \times 10^{-12} F/m$$

    But it, too, will cancel out just like the A's. Your voltage divider expression is nearly correct, but you need to include the voltage being divided and use the supplied values for the distances.
     
  12. Mar 2, 2014 #11
    The supplied voltage (5000v) was a careless mistake from me when typing onto here what I had on my paper.

    I now have VC1 = [itex]\frac{4*\frac{A}{0.001m}}{4*\frac{A}{0.001m}+6*\frac{A}{0.002m}}[/itex] * 5000V = 2857.143 V/m(?)

    Something seems wrong still.

    But assuming this is correct, I now know C and the voltage across C. How do I use this new-found insight to calculate E?
     
  13. Mar 2, 2014 #12

    gneill

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    Yup. The output of a voltage divider is volts, not volts per meter. Voltage is the potential difference across C1.

    What you know is the potential difference in volts and the distance between the plates. Well, you had a formula before that related potential difference to field strength and distance...
     
  14. Mar 2, 2014 #13
    Of course. I blame a messy notebook of formulas.

    V = Ed is the one - hopfully!

    So I have that E = V/d = 2857V/0.001m = ##2.86*10^6## V/m.

    That's hopefully correct. At least I now understand the method to work this out.

    Thanks a lot for not giving up, and thanks a lot for helping me out at all. Really appreciate it.
     
  15. Mar 2, 2014 #14

    gneill

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    Staff: Mentor

    Yes, that looks good.

    No worries, that's why we're here and we're happy to help. Good luck!
     
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