= Solution
Initially the isolated sphere has $Q=4\pi\epsilon_0RV$. After insertion without contact, its charge remains $Q$. In the long-cylinder approximation, with the sphere far from the openings, neglect electric flux escaping through the ends. A <Gauss surface> running through the shielding metal and closed by end sections of negligible flux then encloses zero net charge. Its cavity wall therefore carries $-Q$. Since that cylinder is isolated and initially neutral, its outside wall carries $+Q$. The grounded outer cylinder supplies charge $-Q$ on its inside wall, terminating the annular <electric field lines>.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2018-coaxial-induced-charges.png]
{title=Transverse schematic of a <charged sphere inside a neutral coaxial capacitor>: induced charges and <electric field lines>, with no field inside the conducting walls}
For positive $V$, the cavity <electric field lines> run from the sphere to the inner wall, while annular <electric field lines> run from the outside of the inner conductor to the grounded outer conductor. Negative $V$ reverses their directions. The transverse picture is schematic: exact cavity fields depend on the sphere's position and the ends. The following annular formulas neglect end effects and place the sphere well inside the long cylindrical assembly.
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