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Conductor problem in which the conductor consists of two parallel plates

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    In a vacuum diode, electrons are "boiled" off a hot cathode, at potential zero, and accelerated across a gap to the anode , which is held at positive potential V0. The cloud of moving electrons within the gap(called space charge) quickly builds up to the point where it reduces the field at the surface of the cathode to zero. From then on a steady current I flows between the plate.

    Suppose the plates are large relative to the seperation(A>>d^2), so that edge effects can be neglected. Then V, rho, and v(the speed of the electrons) are all functions of x alone.

    a) Write Poisson's equation for the region between the plates.

    2. Relevant equations
    [tex]\nabla^2[/tex] V= [tex]\nabla[/tex] [tex]\cdot[/tex] E ; If I did not format latex correctly , then here is what I alway trying to write out below:

    del^2 V= del E

    V=[tex]\int[/tex] E [tex]\cdot[/tex] dl


    3. The attempt at a solution

    to get V , I easily integrate E dot dl. How would I obtain the electric field of a parallel plate capacitor. I don't need V to obtain the Poisson equation since:

    [tex]\nabla^2[/tex] V = [tex]\nabla[/tex] [tex]\cdot[/tex] E

    once I calculate E, how would I find the divergence of E? would I set my coordinate system to a cartesian coordinate?
     
    Last edited: Sep 28, 2008
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  3. Sep 28, 2008 #2

    gabbagabbahey

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    I think I recognize this question from Griffith's:smile:. Part (a) is supposed to be very simple:

    Poisson's equation is always:

    [tex] {\nabla}^2V=\frac{\rho}{{\epsilon}_0}[/tex]

    But you're told that [tex]V[/tex] is a function of [tex]x[/tex]...so what is [tex] {\nabla}^2V[/tex]?
     
  4. Sep 28, 2008 #3
    rho=M/V=M/(x^2*d)

    since V=A*d =x^2*d , I am not really sure if the shape of each of the parallel plate is a square , so I can't be sure that A=x^2. If a parallel plate is a square , then rho/epsilon=(M/x^2*d)/epsilon.
     
  5. Sep 28, 2008 #4

    gabbagabbahey

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    It's actually much simpler than that:

    In cartesian coordinates,

    [tex] {\nabla}^2V=\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}[/tex]

    But what are
    [tex]\frac{d^2V}{dy^2}, \quad \frac{d^2V}{dz^2}[/tex]
    if V is a function of x alone?
     
  6. Sep 28, 2008 #5
     
  7. Sep 28, 2008 #6

    gabbagabbahey

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    Yes, part (a) just asks you to write poisson's equation , not solve it.
     
  8. Sep 28, 2008 #7
    I have another question to add:

    b) Assuming the electrons start from rest at the cathode, what is their speed at point x, where x is V(x)?

    W=1/2*CV^2, Work also equals W= 1/2*mv^2

    W(capacitor)= W(Kinetic); therefore wouldn't v=sqrt(CV^2/m)
     
  9. Sep 28, 2008 #8

    gabbagabbahey

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    Yes, but what is the definition of capacitance ,[tex]C[/tex] ?
     
  10. Sep 28, 2008 #9
    C=Q/V ; That means I calculate my velocity in terms of m , V and Q right?
     
  11. Sep 28, 2008 #10

    gabbagabbahey

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    Yes, since C is strictly a geometric property of the plates, and you are not given C, it is better to express v in terms of V, m, q; all quantities that you are either given or can calculate.
     
  12. Sep 29, 2008 #11
    I will post part c and part d of this question:

    part c) In the steady state, I is independent of x. What , then , is the relation between rho an v?

    I use equation for Drift velocity :

    (n*A*vp [tex]\Delta[/tex] t)= [tex]\Delta[/tex]Q not sure how to obtain relation between rho and v.

    part d) Use these three results(part a, b, and c?) to obtain a differential equation for V by eliminating rho and v.

    Not sure where to begin on this part of the problem.
     
  13. Sep 29, 2008 #12

    gabbagabbahey

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    You're sort of on the right track here; What is the definition of current (I'm looking for a very simple differential equation here ;))?
     
  14. Sep 29, 2008 #13
    We really haven't studied current yet; but I remember from my intro physics class that I=dq/dt right?
     
  15. Sep 29, 2008 #14

    gabbagabbahey

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    Right, now what is the relationship between q and rho?
     
  16. Sep 29, 2008 #15

    isn't rho the charge density?
     
  17. Sep 29, 2008 #16

    gabbagabbahey

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    Yes, so what is the mathematical relationship (I'm looking for a simple integral equation)
     
  18. Sep 29, 2008 #17
    J= [tex]\sigma[/tex]*vdrift, J being the current density, [tex]\sigma[/tex] being density of charge per volume.
     
  19. Sep 29, 2008 #18

    gabbagabbahey

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    I was simply looking for this relationship:

    [tex]q=\int_{\mathcal{V}}\rho (\vec{r'}) d\tau'[/tex]

    And since rho is only a function of x in between the plates during the steady state,

    [tex]q= A \int \rho (x) dx[/tex]

    What does this make I=dq/dt?

    Hint: use the chain rule :
    [tex]\frac{dq}{dt}=\frac{dq}{dx} \frac{dx}{dt}[/tex]
     
  20. Sep 29, 2008 #19
    q=A*rho*dx=A*rho*dx

    I=[tex]\delta[/tex] Q/([tex]\delta[/tex] t)

    I=A*[tex]\rho[/tex]*vdrift since, vdrift = dx/dt
     
  21. Sep 29, 2008 #20

    gabbagabbahey

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    Yes,

    [tex]I=\frac{dq}{dt}=\frac{dq}{dx} \frac{dx}{dt}=A \rho (x) v(x)[/tex]

    Now, what do you get for part (d) using this and parts (a) and (b)?
     
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