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Emag - Having trouble following an example

  1. Oct 7, 2006 #1
    I'm having trouble following an example in the book. I don't understand a few steps which I have marked in bold. Any help would be awesome! Thanks.

    Q: In vacuum-tube diodes, electrons are emitted from a hot cathode at zero potential and collected by an anode maintained at a potential [itex] V_0 [/itex], resulting in a convection current flow. Assuming that the cathode and the anode are parallel conducting plates and that hte electrons leave the cathode with a zero initial velocity (spache-charge limited condition), find the relation between teh current density [itex] \vec C [/itex] and [itex] V_0 [/itex].

    A:
    Neglecting fringing effects we have,
    [tex] \vec E(0) = \vec a_y E_y(0) = -\vec a_y \frac{dV(y=0)}{dy} = 0[/tex]

    In the steady state the current density is constant, independent of y:
    [tex] \vec J = -\vec a_y J = \vec a_y \rho(y) u(y) [/tex]
    where the charge density [itex] \rho (y) [/itex] is a negative quanitity. The velocity [itex] \vec u = \vec a_y u(y) [/itex] is related to the electric field intensity [itex] \vec E(y) = \vec a_y E(y) [/itex] by Newton's law of motion:
    [tex] m \vec{d u(y)}{dt} = -eE(y) = e \vec{dV(y)}{dy} [/tex], where [itex] m [/itex] and [itex] e [/itex] are the mass and charge respectively of an electron. Noting that:

    This is where I am confused. I am not noting anything =)

    [tex] m \frac{du}{dt} = m \frac{du}{dy} \frac{dy}{dt} = mu \frac{du}{dy} [/tex]
    [tex] = \frac{d}{dy} \left( \frac{1}{2} mu^2 \right) [/tex]

    [tex] \frac{d}{dy} \left( \frac{1}{2} m u^2 \right) = e \frac{dV}{dy} [/tex]

    I don't understand:
    [tex] m \frac{du}{dy} \frac{dy}{dt} \rightarrow mu \frac{du}{dy} [/tex]
    [tex] mu \frac{du}{dy} \rightarrow \frac{d}{dy} \left( \frac{1}{2} mu^2 \right) [/tex]

    and last but not least...
    [tex] \frac{d}{dy} \left( \frac{1}{2} m u^2 \right) \rightarrow e \frac{dV}{dy} [/tex]
     
    Last edited: Oct 7, 2006
  2. jcsd
  3. Oct 7, 2006 #2

    quasar987

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    This [itex]\vec{a_y}[/itex] buisness is very confusing. Apparently, [itex]\vec{a_y}[/itex] is just a unit vector in the y direction. There's a symbol for that, it's [itex]\vec{j}[/itex] or [itex]\hat{y}[/itex].

    For your first source of confusion:

    According to the equations of kinematics,

    [tex]u(t)=at[/itex]
    [tex]y(t)=\frac{a}{2}t^2[/tex]

    So if you want to write u as a function of y only, it will look like

    [tex]u(y)=\sqrt{2ay}[/tex]

    Now,

    [tex]\frac{du}{dt}=\frac{du}{dy}\frac{dy}{dt}=\frac{du}{dy}\frac{d}{dt}(\frac{a}{2}t^2)=\frac{du}{dy}at=\frac{du}{dy}u[/tex]


    For the second:

    [tex] u \frac{du}{dy} = \frac{d}{dy} \left( \frac{1}{2} u^2 \right) [/tex]

    Just differentiate the RHS to see that it is true.


    Third confusion:

    [itex]\frac{1}{2}mu^2[/itex] is the kinetic energy of an electron. And by conservation of energy, we must have

    [tex]\frac{dK}{dy}=-\frac{dU}{dy}[/tex]

    where U is the potential energy of the electron. But for a charge q, U(y) is also qV(y). Here, q=-e, hence U(y)=-eV(y) and we have

    [tex]\frac{dK}{dy}=e\frac{dV}{dy}[/tex]

    This is what

    [tex] \frac{d}{dy} \left( \frac{1}{2} m u^2 \right) = e \frac{dV}{dy} [/tex]

    is expressing.
     
  4. Oct 8, 2006 #3
    ahhhhhhh..... I feel like slapping myself ;)

    I haven't seen those kinematic equations in awhile, hehe.

    Well this question was an awesome review.

    quasar987 I can't thank you enough. You are always a huge help!
     
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