# Working out the electric field from applied energy

I have created a simulation of one electron bouncing through a 3D mesh of molecules. The electron hopping is determined by a calculation of electron transfer rate using the Marcus equation (a result in units of $1/s$). In order to force the electron to reach one end of this 3D mesh I have applied a driving energy $dE$ (negative) to each molecule.

I am familiar with the electric field equation but I do not see how it applies to one electron bouncing through many molecules. How do I go about calculating the electric field that I applied?

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Then I will expand my question. – Adam Dec 15 '12 at 16:19
Hi Adam, Now it definitely suites the FAQ. Removed comments +1 :-) – Waffle's Crazy Peanut Dec 15 '12 at 17:09

If the energy difference between two sites separated by $\mathbf{r}$, then the effective electric field $\mathbf{E}$ between those two sites is given by $$\mathbf{E}\cdot\mathbf{r}=\frac{dE}{e},$$ where $e$ is the electron charge.

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Alright, so if the energy difference between respective sites is constant dE from molecule to molecule, and there is 1 electron being observed, we use the charge of one electron for dE, the bond length for r and the set dE? – Adam Dec 16 '12 at 11:10
The charge of one electron for $e$, but otherwise, correct. Note that there are a number of important subtleties in interpolating fields from a grid. However, if your grid is uniform and your electric field constant, then this expression will be pretty robust. – KDN Dec 16 '12 at 17:09
I worked it out, but given that the bond length is of course tiny, I'm getting a field of orders of magnitude 10^11. The energy is about 20 milli-electronVolts, which are multiplied by the joule conversion (equal to the charge on the electron) and divided by the charge, to get the quantity as 20 milliJoules/Coulomb - which is then divided by the bond length to get the large value. Does that sound far off? – Adam Dec 17 '12 at 0:45
Yes, that sounds reasonable to me. If your electron gains 20meV at every tiny hop, then $10^{11}$ V/m does not seem like an unreasonable field strength. The solid state density is about $10^{29} \text{m}^{-3}$, or about $10^{10} \text{m}^{-1}$. That's a lot of hops per meter. It seems a little bit high, for a spacing of 1 angstrom, I get a field of $2\times10^8 V/m$. – KDN Dec 17 '12 at 20:35