Coaxial conducting cylinders are located at \(\rho=0.5 \mathrm{~cm}\) and \(\rho=1.2 \mathrm{~cm}\). The region between the cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at \(100 \mathrm{~V}\) and the outer at \(0 \mathrm{~V}\), find (a) the location of the \(20 \mathrm{~V}\) equipotential surface; \((b) E_{\rho \max } ;(c) \epsilon_{r}\) if the charge per meter length on the inner cylinder is \(20 \mathrm{nC} / \mathrm{m}\).

By considering the potentials at the inner and outer cylinders, we can set up the following system of equations: \begin{align*} 100 &= V_{0} + K\ln{\frac{0.5}{\rho_0}} \\ 0 &= V_{0} + K\ln{\frac{1.2}{\rho_0}} \end{align*} By solving this system, we can find the values of \(V_0\) and \(K\).

Solving the second equation for \(V_0\), we get: \[V_{0}=-K\ln{\frac{1.2}{\rho_0}}\] Substitute this into the first equation: \[100 = -K\ln{\frac{1.2}{\rho_0}} + K\ln{\frac{0.5}{\rho_0}}\] Simplifying this equation, we can find the constant \(K\): \[K = \frac{100}{\ln{\frac{0.5}{\rho_0}} - \ln{\frac{1.2}{\rho_0}}} = -50\] Now we can substitute this back into the equation for \(V_0\) to get: \[V_{0} = 50\ln{\frac{1.2}{\rho_0}}\]

To determine the location of the \(20 \mathrm{~V}\) equipotential surface, we need to solve for \(\rho\) in the following equation: \[20 = 50\ln{\frac{1.2}{\rho_0}} -50\ln{\frac{\rho}{\rho_0}}\] By solving this, we can find the location of the equipotential surface \(\rho_{20V}\). In addition, we can find the electric field \(E_{\rho}\) by finding the derivative of the potential function: \[E_{\rho} = -\frac{dV}{d\rho} = \frac{50}{\rho}\] By finding the maximum electric field \(E_{\rho\max}\), we can proceed to step 4 to determine \(\epsilon_{r}\).

To determine the relative permittivity \(\epsilon_{r}\), we need to use Gauss's law and the given charge on the inner cylinder. Gauss's law states that \(\oint \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\epsilon_0}\), and for the coaxial cylinders, it's as follows: \[E_{\rho} 2\pi \rho L = \frac{Q}{\epsilon_0}\] Where \(E_{\rho}\) is the electric field, \(\rho\) is the radial distance, \(L\) is the length of the cylinder, and \(Q\) is the charge per meter length on the inner cylinder. We can plug in the value of \(E_{\rho}\), and the charge \(Q = 20 \mathrm{nC/m}\), and solve for \(\epsilon_{r}\): \[\frac{50}{\rho} 2\pi \rho L = \frac{20 \times 10^{-9}}{\epsilon_0}\] Since \(L\) cancels out, we can find \(\epsilon_{r}\) directly: \[\epsilon_{r} = \frac{50\times 2\pi \times (8.85\times 10^{-12})}{20\times 10^{-9}}\] By calculating this equation, we can now find the value of the relative permittivity \(\epsilon_{r}\).