# Derivation of voltage for vacuum diode space charge region

1. Feb 18, 2009

### Nuahaun

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

$$\frac{d^2V}{dx^2} = constant * V^{-1/2}$$

and I know the solution to be

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

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

$$\int V' dV' = constant*\int V^{-1/2} dV$$

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

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

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. Feb 18, 2009

### gabbagabbahey

I'm not exactly sure what you mean by "separation of variables" in this case. To apply separation of variables, $V$ would have to be a function of more than one variable; but in this case, it is a function of only $x$. 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 $$V^{1/2} d^2 V$$ and $$dx^2$$???

3. Feb 19, 2009

### lanedance

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' = $$\frac{dV}{dx}$$
then
P' = V" = $$\frac{dp}{dx}$$ = $$\frac{dp}{dV}$$.$$\frac{dV}{dx}$$ = p.$$\frac{dp}{dV}$$

otherwise just assume solutions V~xy diff & equate coeffs

Last edited: Feb 19, 2009
4. Feb 19, 2009

### Nuahaun

Thanks for your replies. My mistake was that I though I could solve $V^{1/2} d^2 V = dx^2$, which stemmed from an incomplete understanding of the method of separation of variables (I took the fact that you could solve something like $d^2x/dt^2= F/m$ using double integration, and incorrectly applied that to the differential equation above). I reviewed my notes, and now see why that isn't possible.