Irodov 3.211: The Four Thirds Law of a Diode – Amit & Deepali Sinha Foundation

Problem Irodov 3.211

Two large parallel plates face each other in vacuum. One is a cathode that releases electrons with negligible initial speed, and the electrons stream toward the other plate. Their space charge makes the potential in the gap vary as $\varphi = ax^{4/3}$ , where $a$ is a positive constant and $x$ is the distance from the cathode. Find:

the volume density of the space charge as a function of $x$ ;
the current density.

The two plates in vacuum, the cathode at the bottom, and the distance x measured from it toward the other plate. — The two plates in vacuum, the cathode at the bottom, and the distance $x$ measured from it toward the other plate.

Given

$\varphi = ax^{4/3}$ , $a > 0$
electrons leave the cathode at rest
electron charge $-e$ , mass $m$

Find

(a)

\rho(x)

; (b) the current density

j

Solution

Step 1Space charge from Poisson’s equation

The plates are large, so the potential depends on $x$ alone and Poisson’s equation is one dimensional. Differentiating the given potential twice:

\begin{aligned} \frac{d\varphi}{dx} &= \frac{4}{3}\,a\,x^{1/3}\\ \frac{d^2\varphi}{dx^2} &= \frac{4}{9}\,a\,x^{-2/3} \end{aligned}

Poisson’s equation $d^2\varphi/dx^2 = -\rho/\varepsilon_0$ then gives the charge density:

\rho(x) = -\frac{4}{9}\,\varepsilon_0\, a\, x^{-2/3}

(1)

It is negative, as the charge of the electrons must be, and largest in magnitude next to the cathode, where the electrons are slowest.

The field E_x = -dφ/dx (top) and the potential φ (bottom) across the gap, drawn for a = 1. The field vanishes at the cathode and grows as x^1/3 in magnitude. — The field $E_x = -d\varphi/dx$ (top) and the potential $\varphi$ (bottom) across the gap, drawn for $a = 1$ . The field vanishes at the cathode and grows as $x^{1/3}$ in magnitude.

Step 2The speed of the electrons

An electron starts from rest at the cathode, where $\varphi = 0$ , and gains the kinetic energy $e\varphi$ on reaching the plane $x$ :

\frac{mv^2}{2} = e\varphi = e\,a\,x^{4/3}

Solving for the speed:

v = \sqrt{\frac{2ea}{m}}\;x^{2/3} \tag{2}

Step 3(b) The current density

The current density is the charge density times the velocity of the charge it carries. Multiplying (1) by (2), the powers of $x$ cancel:

j_x = \rho v = -\frac{4}{9}\,\varepsilon_0\,a\,x^{-2/3}\cdot\sqrt{\frac{2ea}{m}}\;x^{2/3}

Its magnitude:

j = \frac{4}{9}\,\varepsilon_0\,a^{3/2}\sqrt{\frac{2e}{m}}

(3)

The electrons move in $+x$ , so the current density points toward the cathode. It is the same in every plane, as it must be in a steady flow with no sources: what enters a slab per second leaves it.

Check

Units: in (1), $a x^{-2/3}$ is a potential over a length squared, so $\varepsilon_0 a x^{-2/3}$ is in $\text{C/m}^3$ ; in (2), $2ea\,x^{4/3}/m$ is twice $e\varphi/m$ , an energy per mass, so its root is a speed; their product is in $\text{A/m}^2$ . Continuity: a current density independent of $x$ is exactly what $\nabla\cdot\vect j = 0$ demands in a steady state, and it comes out only because $\varphi \propto x^{4/3}$ ; any other power would leave a factor of $x$ in (3).

Remark

With $d$ the gap and $V$ the anode voltage, $a = V/d^{4/3}$ turns (3) into the Child law of a vacuum diode, $j = (4/9)\,\varepsilon_0\sqrt{2e/m}\;V^{3/2}/d^2$ . The current is limited by its own space charge, not by the cathode: the field at the cathode is zero, so the electrons there are neither pulled out nor pushed back, and a hotter cathode cannot raise the current at the same voltage.