= Solution
In SI units, <Maxwell's equations> in <vacuum>, allowing <charge density> $\rho$ and <electric current density> $\mathbf J$, are
$$
\boxed{\begin{aligned}
\nabla\cdot\mathbf E&=\rho/\epsilon_0,&\nabla\cdot\mathbf B&=0,\\
\nabla\times\mathbf E&=-\partial_t\mathbf B,&
\nabla\times\mathbf B&=\mu_0\mathbf J+\mu_0\epsilon_0\partial_t\mathbf E.
\end{aligned}}
$$
The source-free specialization sets $\rho=0$ and $\mathbf J=0$; it cannot describe the charged objects at their locations. Applying the <divergence theorem> to the first equation gives <Gauss's law>:
$$
\oint_{\partial\Omega}\mathbf E\cdot d\mathbf S
=\int_\Omega\nabla\cdot\mathbf E\,d^3x
=\frac{Q_{\mathrm{enc}}}{\epsilon_0}.
$$
For an isolated <point charge> $q_1$, <spherical symmetry> makes the <electric field> radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2E_r=q_1/\epsilon_0$. Multiplying by a second <point charge> gives <Coulomb's law>,
$$
\boxed{\mathbf E=\frac{q_1}{4\pi\epsilon_0r^2}\hat{\mathbf r},\qquad
\mathbf F_{2}=\frac{q_1q_2}{4\pi\epsilon_0r^2}\hat{\mathbf r}.}
$$
The <force> points away from the first charge for like signs and toward it for opposite signs.
For the <spherical capacitor>, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi\epsilon_0r^2)$ for $a<r<b$. Integrating the <electric field> inward from the outer shell yields
$$
V(r)-V(b)=-\int_b^r\frac{q}{4\pi\epsilon_0s^2}\,ds
=\frac{q}{4\pi\epsilon_0}\left(\frac1r-\frac1b\right).
$$
An arbitrary constant $V(b)$ remains unless a reference is specified. The <capacitance> is determined by the potential difference:
$$
\boxed{C=\frac{q}{V(a)-V(b)}=4\pi\epsilon_0\frac{ab}{b-a}.}
$$
For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer <voltage> with the outer reference zero. Isolating the inner shell fixes its <electric charge> at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its <electric field> and <electric potential>, now with zero at infinity, are
$$
\mathbf E(r)=\begin{cases}0,&r<a,\\q\hat{\mathbf r}/(4\pi\epsilon_0r^2),&r>a,\end{cases}
\qquad
V(r)=\frac{q}{4\pi\epsilon_0r}\quad(r\geq a).
$$
Consequently
$$
\boxed{V_{\mathrm{surface}}=\frac{bV}{b-a},\qquad
V_{\mathrm{old\ outer\ position}}=\frac{aV}{b-a}.}
$$
These changes of <electric potential> are compatible with constant <electric charge>. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner <electric potential> would instead be $q/(4\pi\epsilon_0a)$.
Finally take the connected spheres to be sufficiently far apart that the long wire's <capacitance> and their mutual influence can be neglected. Their common <electric potential> requires
$$
\frac{q_a}{a}=\frac{q_{2a}}{2a},\qquad q_a+q_{2a}=q,
$$
so $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface <electric fields> are
$$
E_a=\frac{q}{12\pi\epsilon_0a^2},\qquad E_{2a}=\frac{q}{24\pi\epsilon_0a^2},\qquad
\boxed{E_a:E_{2a}=2:1.}
$$
This is the usual widely separated-sphere approximation. A wire enforces an <equipotential>, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net <electric charge> would reside on the exterior and the inner sphere's surface <electric field> would be zero.
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