# Emag - Having trouble following an example

1. Oct 7, 2006

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 $V_0$, 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 $\vec C$ and $V_0$.

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

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

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

$$m \frac{du}{dt} = m \frac{du}{dy} \frac{dy}{dt} = mu \frac{du}{dy}$$
$$= \frac{d}{dy} \left( \frac{1}{2} mu^2 \right)$$

$$\frac{d}{dy} \left( \frac{1}{2} m u^2 \right) = e \frac{dV}{dy}$$

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

and last but not least...
$$\frac{d}{dy} \left( \frac{1}{2} m u^2 \right) \rightarrow e \frac{dV}{dy}$$

Last edited: Oct 7, 2006
2. Oct 7, 2006

### quasar987

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

For your first source of confusion:

According to the equations of kinematics,

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

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

$$u(y)=\sqrt{2ay}$$

Now,

$$\frac{du}{dt}=\frac{du}{dy}\frac{dy}{dt}=\frac{du}{dy}\frac{d}{dt}(\frac{a}{2}t^2)=\frac{du}{dy}at=\frac{du}{dy}u$$

For the second:

$$u \frac{du}{dy} = \frac{d}{dy} \left( \frac{1}{2} u^2 \right)$$

Just differentiate the RHS to see that it is true.

Third confusion:

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

$$\frac{dK}{dy}=-\frac{dU}{dy}$$

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

$$\frac{dK}{dy}=e\frac{dV}{dy}$$

This is what

$$\frac{d}{dy} \left( \frac{1}{2} m u^2 \right) = e \frac{dV}{dy}$$

is expressing.

3. Oct 8, 2006