= Solution
The cylindrical part has length 2 micrometres and <radius> 0.5 micrometres. Including both hemispherical ends gives area $A=2\pi rL+4\pi r^2=3\pi\times10^{-12}$ m$^2$. The <membrane capacitance> is $C=\epsilon_0\epsilon_rA/d\simeq4.17\times10^{-14}$ F. Define the signed inward electrochemical drop per unit positive <Electric charge>[charge] as
$$
\boxed{\Delta p=(\psi_{\mathrm{in}}-\psi_{\mathrm{out}})-\frac{2.303k_BT}{e}(\mathrm{pH}_{\mathrm{in}}-\mathrm{pH}_{\mathrm{out}}).}
$$
There is a factor-of-two ambiguity in the wording of the <Electric charge>[charge] data. If the stated excess means the inner <capacitor> <Electric charge>[charge] $Q=-40000e$, then $\Delta\psi=Q/C=-0.154$ V and $\Delta p\simeq-0.183$ V at 298 K. If “inside minus outside” is taken literally as $Q_{\mathrm{in}}-Q_{\mathrm{out}}=-40000e$ for opposite <capacitor> <Electric charge>[charges], the inner plate has $-20000e$, giving $\Delta\psi=-0.0768$ V and $\Delta p\simeq-0.106$ V. The latter is the literal charge-difference interpretation; the former is the commonly intended excess-charge convention. In either case inward <proton> motion is favorable, and the positive available driving magnitude is $-\Delta p$.
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