1. Problem statement: A solid conducting sphere of radius $a$ has net charge $2Q$. A concentric conducting spherical shell with inner radius $b$ and outer radius $c$ has net charge $-Q$. The spheres are connected by a wire for an instant. We need to find the surface charge densities on the inner and outer surfaces of the spherical shell. 2. Physical principle: When two conductors are connected by a wire, charges redistribute to equalize the potential on both conductors. 3. Since the shell is a conductor, the electric field inside its material must be zero. This forces all excess charges on the shell to reside only on the surfaces. 4. Because the shell initially has charge $-Q$, and the inner sphere has charge $2Q$, when connected by a wire, charge flows until both are at the same potential. 5. To maintain zero field inside the shell material, the inner surface of the shell must have charge $-2Q$ (opposite the solid sphere's charge of $2Q$). 6. The shell has total net charge $-Q$, so the outer surface charge must be $(-Q) - (-2Q) = Q$. 7. Surface charge density $\sigma = \frac{Q}{4 \pi r^2}$ for sphere of radius $r$. 8. Hence the surface charge density on inner shell surface (radius $b$): $$\sigma_{inner} = \frac{-2Q}{4 \pi b^2}$$ 9. Surface charge density on outer shell surface (radius $c$): $$\sigma_{outer} = \frac{Q}{4 \pi c^2}$$ 10. The provided answer (c) corresponds to inner surface charge density 0, which doesn't align with the physics above; the correct answer is as in step 8 and 9. Final answer: Inner surface: $$\sigma_{inner} = \frac{-2Q}{4 \pi b^2}$$ Outer surface: $$\sigma_{outer} = \frac{Q}{4 \pi c^2}$$