 Solution

The cylindrical part has length 2 micrometres and radius 0.5 micrometres. Including both hemispherical ends gives area $A=2\pi rL+4\pi r^2=3\pi \times 10^{-12}$ m${}^2$. The membrane capacitance is $C={ϵ}_0{ϵ}_{r}A/d\simeq 4.17\times 10^{-14}$ F. Define the signed inward electrochemical drop per unit positive charge asThere is a factor-of-two ambiguity in the wording of the charge data. If the stated excess means the inner capacitor 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 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 $-\Delta p$.