= Solution
The positive charge per unit interface area on the n side and negative charge on the p side are
$$
Q_n=\int_0^{w/2}e\alpha x\,dx=\frac{e\alpha w^2}{8},\qquad
Q_p=-Q_n.
$$
Use the signed p-side charge $Q=Q_p$ because $V_{\rm ext}$ is the p-side voltage relative to n. Differentiating the width relation gives
$$
\frac{dw}{dV_{\rm ext}}=-\frac{w}{3(V_b-V_{\rm ext})},\qquad
C=\frac{dQ_p}{dV_{\rm ext}}
=\frac{e\alpha w^2}{12(V_b-V_{\rm ext})}.
$$
Using part (e), the positive <junction capacitance> per unit area is
$$
\boxed{C=\frac{\varepsilon_s}{w}
=\left[\frac{e\alpha\varepsilon_s^2}{12(V_b-V_{\rm ext})}\right]^{1/3}.}
$$
It decreases with increasing reverse-bias magnitude as $(V_b+|V_{\rm ext}|)^{-1/3}$. Equivalently,
$$
\boxed{C^{-3}=\frac{12(V_b-V_{\rm ext})}{e\alpha\varepsilon_s^2},}
$$
so a linear $C^{-3}$ versus voltage plot diagnoses this ideal linear grading and can determine $\alpha$.
For this <linearly graded junction capacitance>, the charge convention matters: if “stored charge” means the positive magnitude $Q_n$, then $dQ_n/dV_{\rm ext}=-\varepsilon_s/w$; the positive measured capacitance is $-dQ_n/dV_{\rm ext}=dQ_n/dV_R$ for $V_R=-V_{\rm ext}$. The total net depletion charge is zero and is not the stored terminal charge. At substantial forward bias, <minority carrier> storage and the failure of the sharp <depletion approximation> prevent extrapolation of the apparent divergence at $V_{\rm ext}=V_b$.
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