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Energy of different capacitor

  1. Apr 22, 2014 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate capacitor with plates of area (0.5m) * (1m) has a distance separation of 2 [cm] and a voltage difference of V = 200 [V], as shown in Fig.

    figura.png

    a) Find the energy stored

    b) keep d1 = 2 [cm] and the voltage difference V, while increasing d2 = 2.2 [cm]. Find the energy stored (hint hint u=1/2CV^2)

    2. Relevant equations

    ecuaciones_usadas.png

    3. The attempt at a solution

    intento.png
     
  2. jcsd
  3. Apr 22, 2014 #2

    Simon Bridge

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    You could try dividing the capacitor into lots of narrow capacitors in parallel.
     
  4. Apr 22, 2014 #3
    but how if d1=2 and d2=2.2, can you still use the capacitor formulas even if its a uneven capacitor?
     
  5. Apr 22, 2014 #4

    Simon Bridge

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    You have to divide it into smaller capacitors than that.
     
  6. Apr 22, 2014 #5
    can you give me a little example please, i just cannot imagine since one side is uneven :(
     
  7. Apr 23, 2014 #6

    Simon Bridge

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    Have you not done any calculus?
     
  8. Apr 23, 2014 #7
    so for example i will use this formula E = Δ V / Δ d, and by dividing the plates in very tiny plates
    i can now write E as E = ∑ (Δ V / Δd), but since we V is constant E= V (∑ (1/Δd)) so this will be the same as E=V∫((1/dx)) from d1 to d2 where d=x???

    is this approach correct? and then calculate the charge and then the energy????
     
  9. Apr 23, 2014 #8
    ok so i was wondering if so for example i will use this formula E = Δ V / Δ d, and by dividing the plates in very tiny plates i can now write E as E = ∑ (Δ V / Δd), but since we V is constant E= V (∑ (1/Δd)) so this will be the same as E=V∫((1/dx)) from d1 to d2 where d=x???

    is this approach correct? and then calculate the charge and then the energy????
     
  10. Apr 23, 2014 #9

    rude man

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    Looks unwieldy. What happens to your stack of plates when you reach d = 2cm?

    How about this instead: compute the E field as a function of x along the bottom plate: x = 0 on the left and x = 0.5m on the right. Compute the energy in a typical volume dx d(x) 1m and integrate.
     
  11. Apr 23, 2014 #10
    the idea is that i can divide the plates in very tine ones and will use this forumula

    [tex] Q \equiv εA(Δ V / Δd) [/tex] and will set [tex]A\equiv dxdy[/tex] and [tex]d \equiv dz[/tex]

    so the new equation will be like
    [tex] Q \equiv εV \int_0^1 {dx} \int_0^.5 {dy} \int_a^b{1/dz}[/tex]

    where [tex]a\equiv2[/tex] and [tex]b\equiv2.2[/tex]


    and then use [tex]u \equiv QV(1/2)[/tex] is this correct????
     
  12. Apr 23, 2014 #11

    Simon Bridge

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    Kinda... why introduce the charge?

    You want to use ##U=\frac{1}{2}CV^2\implies dU = \frac{1}{2}V^2\;dC##

    Put the x-axis in the direction the separation of the plates varies.
    Divide the width into strips L=1m long, and dx wide... this is a row of parallel plate capacitors in parallel... you know how to find the capacitance of a parallel plate capacitor.

    If we put the x axis so the separation goes like: d(x=0)=2m and d(x=0.5)=2.2m, you can find an expression for d(x)

    From that, and knowing that V is the same for all the capacitor elements, you can find:
    dC= (the capacitance of the element at position x)
    ... get this from the expression for the capacitance of a parallel plate capacitor.

    Use that to get an expression of form: ##dU = f(x)\;dx##, and integrate both sides.
     
  13. Apr 23, 2014 #12

    rude man

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    There's no need to compute capacitance, though it can be done that way of course.

    The direct way is to use the formula energy = energy density times volume. What is the energy density of a field E in vacuo?
     
  14. Apr 23, 2014 #13
    using [tex] C\equiv \frac{εA}{d} [/tex] where A= area of plates and the distance is a function of x [tex] d(x) \equiv 2 + 0.4x[/tex]

    we get

    [tex] C\equiv ε \frac{\int_{0}^{1}dy \int_{0}^{.5}dx}{\int_{0}^{.5}(2+.4x)dx} [/tex]


    then substituting in [tex] U \equiv \frac{1}{2} ε V^2 C [/tex]

    we can get

    [tex] U \equiv \frac{1}{2} ε V^2 \frac{\int_{0}^{1}dy \int_{0}^{.5}dx}{\int_{0}^{.5}(2+.4x)dx} [/tex]

    is this correct?
     
    Last edited: Apr 23, 2014
  15. Apr 23, 2014 #14
    is this approach correct in my last reply?
     
  16. Apr 23, 2014 #15

    Simon Bridge

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    I don't think so - you have too many integrals.
     
  17. Apr 23, 2014 #16
    where am i wrong?
     
  18. Apr 23, 2014 #17

    Simon Bridge

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    Why not go step-by-step through the suggestion in post #11?

    you have:
    1. expression for d(x): ##d=2+2x/5## (better to avoid decimals in equations - if you can)

    You still need the others:
    If the plates are 1m long and dx wide - what is their area?

    Therefore:
    2. dC=

    Therefore:
    3. dU=
     
  19. Apr 23, 2014 #18
    ok so after trying with my friend we got this,

    image.png

    is this right now?
     
  20. Apr 23, 2014 #19

    Simon Bridge

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    Since you will not answer questions or follow suggestions, I cannot help you.
     
  21. Apr 23, 2014 #20
    is the last reply right?
     
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