# Field emission vs. electrical breakdown

In vacuum a tungsten needle sits in front of a copper plate at some separation $d$. How does the ratio of the voltage at a fix field emission current (e.g. 1 pA) $V_{fe}(1 \ \text{pA})$ and the breakdown voltage $V_{br}$ scale with the distance $d$?

$\frac{V_{fe}(1 \ \text{pA})}{V_{br}} \propto d^\alpha$

Intuitively I think this ratio decreases for increasing $d$, but are there any formulas or at least arguments to justify it?

Field emission is given by the (adjusted) Fowler-Nordheim formula. We obtain for a fixed curent $I \propto E_s^2 e^{-k E_s},$ with $k$ depending on the work function, $E_s$ being the local electric field at the surface.
Breakdown occurs when the local electric field reaches a value $E_{br}$.
In this simple scenario of two isolated electrodes we can assume $V \propto E \; d.$ Since both phenomena depend directly on the electric field, only the latter expression is relevant. Thus the ratio $\frac{V_{fe}(I)}{V_{br}} \propto 1$.