= Full-depletion capacitance of a symmetric junction slab
{title2=$C_{\rm fd}=\varepsilon_s\mathcal A/(2d)$}
Let each side of a symmetric abrupt <p-n junction> have thickness $d$, area $\mathcal A$ and doping density $N$. Partial depletion gives $w^2=\varepsilon_s(V_{\rm bi}-V)/(eN)$. Both sides become fully depleted at $V_{\rm fd}=V_{\rm bi}-eNd^2/\varepsilon_s$. Beyond that point, the ionized dopant charges are fixed, but an incremental terminal voltage changes the electric field uniformly across the full thickness $2d$. <Gauss's law> then gives $dQ_{\rm terminal}=\varepsilon_s\mathcal A\,dV/(2d)$, up to terminal orientation. The measured positive capacitance approaches $C_{\rm fd}$ rather than vanishing. This geometrical result assumes ideal contacts, negligible mobile carriers, no breakdown and no other series capacitances.
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