I have a thin film with known dimensions and conductivity. If I hook up wires and apply a bias, how can I calculate/estimate the current that I will get? Here is where I'm at:

$$ \mathbf{J}=\sigma \mathbf{E} $$ For a constant electric field, $E=V/d$, so $$J=\frac{\sigma V}{d}$$ So this gives me J in the film. I understand that $E=\int_S \bf{J}\cdot \, \mathrm{d}\mathbf{A}$.

My question is: what cross-sectional area do I use for the integral? Should it be for the wire I plan to connect?

  • $\begingroup$ The electric field within the thin film will depend on the shape, resistivity, and locations of the electrodes. For example, if the two electrodes are just point contacts, the current density field and the electric field will look like the electric field you would expect around positive and negative charges placed at the electrode positions. If the electrodes have very low resistivity and extend along two opposite edges of a rectangular film, the electric field lines and current density vectors will be essentially uniform and parallel, running from one electrode to the other. $\endgroup$
    – S. McGrew
    Jan 1 '19 at 17:58
  • $\begingroup$ Since $\mathbf{J}$ is zero outside of the conductor, any cross-section entirely encircling it will do. But I am not following you where you state that $\mathbf{E}$ is constant. Also, I'm not clear, are you are asking about the total current, or the current density distribution within the film? $\endgroup$
    – kkm
    Jan 2 '19 at 6:12
  • $\begingroup$ Adding to previous comments, please make it clearer what you are asking. You could start by explicitly naming the quantities that you use, for example V, d and E. You wrote "E = integral over S of J * dA". The right side describes the current through area S, so I would have expected "I" instead of "E" on the left side here. Is the current in question flowing mainly in the film plane, or effectively perpendicular to the film plane (top to bottom for example)? $\endgroup$ Jan 2 '19 at 14:34

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