a
We suppose that the cavity is filled up by a positive as well as negative volume charge of \(\rho .\) So the electric field now produced at \(P\) is the superposition of two electric fields.

\((i)\) The electric field created due to the infinitely long solid cylinder is

\(E_{1}=\frac{\rho R}{4 \varepsilon_{0}}\) directed towards the \(+Y\) direction

\((ii)\) The electric field created due to the spherical negative charge density

\(\mathrm{E}_{2}=\frac{\rho \mathrm{R}}{96 \varepsilon_{0}}\) directed towards the \(-Y\) direction.

\(\therefore \) The net electric field is

\(\mathrm{E}=\mathrm{E}_{1}-\mathrm{E}_{2}=\frac{1}{6}\left[\frac{23 \rho \mathrm{R}}{16 \varepsilon_{0}}\right]\)