1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivation of voltage for vacuum diode space charge region

  1. Feb 18, 2009 #1
    I have found the 1D differential equation relating voltage and position for a vacuum diode in the space charge region, which is

    [tex] \frac{d^2V}{dx^2} = constant * V^{-1/2} [/tex]

    and I know the solution to be

    [tex] V(x) = V_0 \left(\frac{x}{d}\right)^{4/3} [/tex]

    which is found by multiplying both sides by [itex] V' = \frac{dV}{dx} [/itex] and then integrating the following expression with homogeneous boundary conditions:

    [tex] \int V' dV' = constant*\int V^{-1/2} dV [/tex]

    What I don't understand is why this trick is even necessary. As far as I can tell the differential equation can be solved by a simple separation of variables, which gives an answer of

    [tex] V(x) = V_0 \left(\frac{x}{d}\right)^{4/5} [/tex]

    The two answers are different so obviously it's a mistake to separate variables, but for the life of me I can't tell where it is. Could anyone enlighten me? Thanks.
  2. jcsd
  3. Feb 18, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    I'm not exactly sure what you mean by "separation of variables" in this case. To apply separation of variables, [itex]V[/itex] would have to be a function of more than one variable; but in this case, it is a function of only [itex]x[/itex]. SOV is typically used to turn PDEs into ODEs, not to solve an ODE.

    I can only guess that what you meant to say was that you tried treating it as a "separable differential equation"; but you have a second order DE....how exactly does one integrate [tex]V^{1/2} d^2 V[/tex] and [tex]dx^2[/tex]???
  4. Feb 19, 2009 #3


    User Avatar
    Homework Helper

    yeah wasn't totally sure what was meant, however a trick to reduce it to 2 first order DEs as follows:

    First write
    P = V' = [tex]\frac{dV}{dx}[/tex]
    P' = V" = [tex]\frac{dp}{dx}[/tex] = [tex]\frac{dp}{dV}[/tex].[tex]\frac{dV}{dx}[/tex] = p.[tex]\frac{dp}{dV}[/tex]

    otherwise just assume solutions V~xy diff & equate coeffs
    Last edited: Feb 19, 2009
  5. Feb 19, 2009 #4
    Thanks for your replies. My mistake was that I though I could solve [itex] V^{1/2} d^2 V = dx^2 [/itex], which stemmed from an incomplete understanding of the method of separation of variables (I took the fact that you could solve something like [itex] d^2x/dt^2= F/m [/itex] using double integration, and incorrectly applied that to the differential equation above). I reviewed my notes, and now see why that isn't possible.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook