A thin, hollow, metal cylinder of radius \(R\) has a surface charge distribution \(\sigma\). A long, thin wire with a linear charge density \(\lambda / 2\) runs through the center of the cylinder. Find an expression for the electric fields and the direction of the field at each of the following locations: a) \(r \leq R\) b) \(r \geq R\)

In this exercise, you are asked to find the electric fields inside and outside a metal cylinder with a surface charge distribution and a linear charge density. To solve this problem, follow these steps: 1. Determine the electric field due to the surface charge distribution of the cylinder. Inside the metal cylinder (r ≤ R), the electric field due to the surface charge distribution is zero. Outside the metal cylinder (r ≥ R), the electric field can be found using Gauss's Law: \(E_\sigma = \frac{\sigma R}{\varepsilon_0r}\). 2. Determine the electric field due to the linear charge density of the wire. Using Gauss's Law, the electric field can be found for both inside and outside the metal cylinder (r ≤ R and r ≥ R): \(E_\lambda = \frac{\lambda}{4\pi\varepsilon_0 r}\). 3. Determine the electric field at the given locations by superposition. a) For r ≤ R: \(E = E_\lambda = \frac{\lambda}{4\pi\varepsilon_0 r}\). b) For r ≥ R: \(E = E_\sigma + E_\lambda = \frac{\sigma R}{\varepsilon_0r} + \frac{\lambda}{4\pi\varepsilon_0 r}\). 4. Determine the direction of the electric field. In both cases, the electric field is radially outward from the wire. Based on these results, the electric field inside the metal cylinder (r ≤ R) is given by \(E = \frac{\lambda}{4\pi\varepsilon_0 r}\) and is directed radially outward from the wire. Outside the metal cylinder (r ≥ R), the electric field is given by \(E = \frac{\sigma R}{\varepsilon_0r} + \frac{\lambda}{4\pi\varepsilon_0 r}\) and is also directed radially outward from the wire.