= Photoelectric contact-potential correction
{title2=$V_s=V_{\rm app,stop}+V_c$}
Define $V_c$ as positive when the contact field adds to the retarding barrier, and let $V_{\rm app}$ be the signed externally imposed retarding voltage. In this chosen sign convention the barrier is $e(V_{\rm app}+V_c)$, hence
$$
V_{\rm app,stop}=\frac{h}{e}\nu-\frac{\Phi}{e}-V_c,
\qquad V_s=V_{\rm app,stop}+V_c.
$$
This convention specifies a model of the gap; an instrument's polarity labels must be translated into it. With a fixed $V_c$, subtraction at two <frequencies> removes both the <work function> and the offset:
$$
\frac{V_{\rm app,stop}(\nu_2)-V_{\rm app,stop}(\nu_1)}{\nu_2-\nu_1}=\frac{h}{e}.
$$
For an original example with the same $\Phi=2.30\,\mathrm{eV}$ and an additional $V_c=+0.200\,\mathrm V$, choose $\nu_1=6.00\times10^{14}\,\mathrm{Hz}$ and $\nu_2=1.00\times10^{15}\,\mathrm{Hz}$. The predicted terminal endpoints are $-0.018599$ and $1.635668\,\mathrm V$. The negative first endpoint calls for a small externally accelerating voltage because the contact field already over-retards the fastest <Electrons>. Their difference is $1.654267\,\mathrm V$, giving $h/e\simeq4.135668\times10^{-15}\,\mathrm{V\,s}$. Ignoring $V_c$ in the intercept would falsely infer $2.50\,\mathrm{eV}$ for the <work function>. The terminal line's zero would then be mistaken for a material threshold. If $V_c$ drifts between the two readings, even the slope is biased. These are calculated examples, not Millikan's measurements.
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