 Solution

Initially the isolated sphere has $Q=4\pi {ϵ}_{0}RV$. 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.

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.