A coaxial cable consists of a wire of radius \(a\) surrounded by a thin metal cylindrical shell of radius \(b\). The wire has a uniform linear charge density \(\lambda>0\) and the outer shell has a uniform linear charge density \(-\lambda\). (a) Sketch the field lines for this cable. (b) Find expressions for the magnitude of the electric field in the regions \(r \leq a, a<r<b,\) and \(b \leq r\)

Draw a coaxial cable, showing a positively charged wire of radius \(a\) and a negatively charged cylindrical shell of radius \(b\). The electric field lines should originate from the positive charge on the wire and terminate on the negative charge of the outer shell. Inside the central wire, the electric field lines should be straight and radial, between the wire and the shell, the lines should be curved and directed radially outward from the wire, and beyond the shell, the electric field lines should be continuous and parallel to the cable axis.

Consider a cylindrical Gaussian surface with radius \(r\) and length \(L\) such that \(0 \leq r \leq a\). According to Gauss's Law, the electric flux through this surface is equal to the total charge enclosed by the surface divided by the vacuum permittivity \(\epsilon_0\): \(\Phi_E = \frac{Q_\text{enclosed}}{\epsilon_0}\). The total charge enclosed by this surface is given by the product of the linear charge density \(\lambda\) and the length \(L\), that is \(Q_\text{enclosed} = \lambda L\). The electric field inside the wire is radial and uniform, so the electric flux is simply the product of the field strength \(E\), the area of the surface, and the cosine of the angle between the field and the normal to the surface (\(\cos{0} = 1\)). Therefore, \(\Phi_E = EA_\text{cylindrical} = E (2\pi r L)\). According to Gauss's Law, \(\Phi_E = \frac{\lambda L}{\epsilon_0}\), which leads to the expression for the electric field inside the wire: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\) for \(0 \leq r \leq a\).

Now, consider a cylindrical Gaussian surface with radius \(r\) and length \(L\) such that \(a < r < b\). For this case, the total charge enclosed by the surface is still given by the product of the linear charge density \(\lambda\) and the length \(L\), that is \(Q_\text{enclosed} = \lambda L\). The electric field in the region between the wire and the shell is still radial and uniform, so following the same steps as before, we find the expression for the electric field: \(E = \frac{\lambda}{2\pi\epsilon_0 r}\) for \(a < r < b\).

Lastly, consider a cylindrical Gaussian surface with radius \(r\) and length \(L\) such that \(r \geq b\). In this case, the total charge enclosed by the surface includes both the charge on the wire and the charge on the shell. Since their magnitudes are equal and opposite, the net charge enclosed is zero: \(Q_\text{enclosed} = 0\). According to Gauss's Law, this implies that the electric field outside the shell is zero: \(E = 0\) for \(r \geq b\). Thus, the expressions for the magnitude of the electric field in the three regions are: 1) Inside the wire (\(0 \leq r \leq a\)): \(E = \frac{\lambda}{2\pi\epsilon_0 r}\). 2) Between the wire and the shell (\(a < r < b\)): \(E = \frac{\lambda}{2\pi\epsilon_0 r}\). 3) Outside the shell (\(r \geq b\)): \(E = 0\).