= Solution
Including both flat ends of the cylindrical cell, its membrane <surface area> is
$$
A=2\pi r\ell+2\pi r^2
=2\pi(10^{-6})(100\times10^{-6})+2\pi(10^{-6})^2
=6.34602\times10^{-10}\,\mathrm{m^2}.
$$
The <capacitance> is $C=(0.01\,\mathrm{F\,m^{-2}})A=6.34602\,\mathrm{pF}$. The <electric charge> needed for the specified <depolarization> is therefore
$$
\Delta Q=C\Delta\Psi=5.39411\times10^{-13}\,\mathrm C,\qquad
\boxed{N=\frac{|\Delta Q|}{e}=3.36674\times10^6\text{ monovalent ions}.}
$$
Ignoring the ends gives $3.33\times10^6$, 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 $\pi r^2\ell=3.14159\times10^{-16}\,\mathrm{m^3}$. Distributed over this <volume>, the transferred number would correspond to only $0.0177955\,\mathrm{mM}$. Relative to an \b[illustrative], not source-specified, $100\,\mathrm{mM}$ internal ionic pool, that is a fraction $1.78\times10^{-4}$. 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.
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