Let the cylindrical surfaces \(\rho=4 \mathrm{~cm}\) and \(\rho=9 \mathrm{~cm}\) enclose two wedges of perfect dielectrics, \(\epsilon_{r 1}=2\) for \(0<\phi<\pi / 2\) and \(\epsilon_{r 2}=5\) for \(\pi / 2<\phi<2 \pi\) If \(\mathbf{E}_{1}=(2000 / \rho) \mathbf{a}_{p} \mathrm{~V} / \mathrm{m}\), find \((a) \mathbf{E}_{2} ;(b)\) the total electrostatic energy stored in a \(1 \mathrm{~m}\) length of each region.

We are given \(\mathbf{E}_{1}=(2000 / \rho) \mathbf{a}_{p} \mathrm{~V} / \mathrm{m}\) in the first region with \(\epsilon_{r 1}=2\). To find \(\mathbf{E}_{2}\), we need to use the boundary conditions for the electric field at the interface, which states that the tangential component of the electric field is continuous across the interface. Since the electric field given for region 1 has only a radially outward component, there is no tangential component at the interface. Hence, there will be no tangential component in region 2 as well. So, we can assume that the electric field in region 2 has only a radial component, i.e., \(\mathbf{E}_{2} = E_{2} \mathbf{a}_{p}\). Now, we can use the boundary condition for the normal component of the electric field which states that: \(\epsilon_{1} E_{1n} = \epsilon_{2} E_{2n}\) \(\epsilon_{r 1} \epsilon_{0} E_{1} = \epsilon_{r 2} \epsilon_{0} E_{2}\) Solving for \(E_{2}\): \(E_2 = \frac{\epsilon_{r 1}}{\epsilon_{r 2}} E_1\) \(E_2 = \frac{2}{5} 2000 / \rho\) \(\mathbf{E}_{2} = (\frac{2}{5} \frac{2000}{\rho}) \mathbf{a}_{p} \mathrm{~V} / \mathrm{m}\)

We can now calculate the electrostatic energy density in each region using the formula: \(u = \frac{1}{2} \epsilon E^2\) So, for region 1: \(u_{1} = \frac{1}{2} \epsilon_{r 1} \epsilon_{0} E_{1}^2 = \frac{1}{2} \cdot 2 \epsilon_{0} \cdot \frac{2000^2}{\rho^2}\) For region 2: \(u_{2} = \frac{1}{2} \epsilon_{r 2} \epsilon_{0} E_{2}^2 = \frac{1}{2} \cdot 5 \epsilon_{0} \cdot (\frac{2}{5} \frac{2000}{\rho})^2\) To find the total electrostatic energy stored in a 1m length of each region, we integrate the energy density over the volume (considering a radial distance of \(\rho\) for 1m length in each region): \(W_{1} = \int_{0}^{\pi/2} \int_{4\text{cm}}^{9\text{cm}} u_{1} \rho \, d\rho \, d\phi\) \(W_{1} = \int_{0}^{\pi/2} \int_{4\text{cm}}^{9\text{cm}} \frac{1}{2} \cdot 2 \epsilon_{0} \cdot \frac{2000^2}{\rho^2} \rho \, d\rho \, d\phi\) \(W_{1} = \epsilon_{0} 2000^2 \int_{0}^{\pi/2} d\phi \int_{4\text{cm}}^{9\text{cm}}\frac{1}{\rho} d\rho\) Similarly, for region 2: \(W_{2} = \int_{\pi/2}^{2\pi} \int_{4\text{cm}}^{9\text{cm}} u_{2} \rho \, d\rho \, d\phi\) \(W_{2} = \int_{\pi/2}^{2\pi} \int_{4\text{cm}}^{9\text{cm}} \frac{1}{2} \cdot 5 \epsilon_{0} \cdot (\frac{2}{5} \frac{2000}{\rho})^2 \rho \, d\rho \, d\phi\) \(W_{2} = 2\epsilon_{0} 2000^2 \int_{\pi/2}^{2\pi} d\phi \int_{4\text{cm}}^{9\text{cm}}\frac{1}{\rho} d\rho\) Now we can calculate the values of \(W_{1}\) and \(W_{2}\): \(W_{1} = \epsilon_{0} 2000^2 (\frac{\pi}{2} - 0)(\ln(9\text{cm}) - \ln(4\text{cm}))\) \(W_{2} = 2\epsilon_{0} 2000^2 (2\pi - \frac{\pi}{2})(\ln(9\text{cm}) - \ln(4\text{cm}))\) So, the total electrostatic energy stored in a 1m length of each region is given by \(W_{1}\) and \(W_{2}\).