Final answer:
To find the electric field and surface charge densities, we utilize Gauss's Law and consider the properties and dimensions of a charged rod and a coaxial cylindrical shell.
Step-by-step explanation:
The student asks about the magnitude of the electric field at a distance from a coaxial cylindrical shell and the surface charge density on the inner and outer surfaces of the shell with a charge of uniform linear density on a long, thin rod at the axis.
In this case, we apply Gauss's Law to calculate the electric field. Since the rod has a linear charge density of 2.0 nC/m, we can say λ = 2.0 × 10^{-9} C/m. For a cylindrical Gaussian surface with a radius of 15 cm, which encloses the charged rod, the electric field E at that surface can be calculated using the equation E × (2π× r × L) = λ× L / ε_{0}, where L is the length of the cylinder, r is the radius, and ε_{0} is the vacuum permittivity (8.854 × 10^{-12} C^{2}/N·m^{2}). We find that E = λ / (2πε_{0}r).
Next, for the surface charge densities on the inner and outer surface of the shell, we understand that the inner surface must have a charge density equal in magnitude but opposite in sign to the linear charge density of the rod, to preserve the net charge of zero on the shell. Hence, σ_{inner} = -λ / (2π × inner radius). The outer surface must have an equal and opposite charge to maintain the net charge of zero, which is σ_{outer} = -σ_{inner}. We then calculate the values using the given dimensions.