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.(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Homework Help: Emag - Having trouble following an example

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