Including both flat ends of the cylindrical cell, its membrane surface area is

A = 2 π r ℓ + 2 π r^{2} = 2 π (1 0^{- 6}) (100 \times 1 0^{- 6}) + 2 π (1 0^{- 6})^{2} = 6.34602 \times 1 0^{- 10} m^{2} .

The capacitance is

C = (0.01 F m^{- 2}) A = 6.34602 pF

. The electric charge needed for the specified depolarization is therefore

Δ Q = C ΔΨ = 5.39411 \times 1 0^{- 13} C, N = \frac{∣Δ Q ∣}{e} = 3.36674 \times 1 0^{6} monovalent ions .

(2)

Ignoring the ends gives

3.33 \times 1 0^{6}

, only about one percent smaller. Inward movement of positive ions, or outward movement of negative ions, produces the stated positive change in the inside-minus-outside membrane potential.

The cell volume is

π r^{2} ℓ = 3.14159 \times 1 0^{- 16} m^{3}

. Distributed over this volume, the transferred number would correspond to only

0.0177955 mM

. Relative to an illustrative, not source-specified,

100 mM

internal ionic pool, that is a fraction

1.78 \times 1 0^{- 4}

. Thus a substantial change in membrane potential requires movement of only a tiny fraction of bulk ions; it does not require wholesale depletion of the intracellular solution. The separated electric charge is concentrated near the two membrane surfaces.