= Solution
Each addition starts a new statistically identical waiting episode at zero gap. Thus the addition rate is the reciprocal of a <mean first-passage time>, not the reciprocal of the mean gap:
$$
\boxed{\tau=\frac1{j_0}=\frac{e^{\beta F\delta}-\beta F\delta-1}{D\beta^2F^2}.}
$$
This can also be derived without the steady density. If $T(x)$ is the mean time to reach $\delta$ starting at gap $x$, its backward <Smoluchowski equation> is
$$
DT''-D\beta FT'=-1,\qquad T'(0)=0,\qquad T(\delta)=0.
$$
The first boundary is reflecting and the second absorbing. An <integrating factor> gives $T'(x)=[1-e^{\beta Fx}]/(D\beta F)$, whence
$$
T(0)=-\int_0^\delta T'(x)\,dx
=\frac{e^{\beta F\delta}-1-\beta F\delta}{D\beta^2F^2}=\tau.
$$
At weak <force>, $\tau\simeq\delta^2/(2D)$ is the time for a freely diffusing wall to move a <monomer> length, with reflection at the tip. At strong <force>, opening the gap requires <work> $F\delta$, with <Boltzmann distribution> suppression $e^{-\beta F\delta}$. The drift establishes a near-tip thermal length $\ell_F=k_BT/F$ and a <diffusion> time $\ell_F^2/D$; consequently
$$
\tau\sim\frac{\ell_F^2}{D}e^{\delta/\ell_F}.
$$
The exponentially long wait explains why the growth slows. There is no additional chemical waiting time in the imposed absorbing-threshold model; such a time would be needed to describe the separate <reaction-limited polymerization Brownian ratchet> regime.
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