= Solution
Spherical symmetry and the vacuum <Gauss's law> imply $\mathbf E=A\hat{\mathbf r}/r^2$ in the gap, where $A$ is constant. Integrating the <electric field> between the two <equipotentials> gives
$$
V_1=V(a)-V(b)=\int_a^b\frac{A}{r^2}\,dr=A\left(\frac1a-\frac1b\right).
$$
Consequently
$$
\boxed{\mathbf E(r)=\frac{V_1ab}{(b-a)r^2}\hat{\mathbf r},\qquad a<r<b}.
$$
The inner charge is $Q=4\pi\epsilon_0A$, which reproduces the <spherical capacitor> <capacitance> $4\pi\epsilon_0ab/(b-a)$. The radial direction is outward for positive $V_1$ and inward for negative $V_1$; the formula represents the signed field, not just its magnitude.
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