A line charge of infinite length lies along the \(z\) axis and carries a uniform linear charge density of \(\rho_{\ell} \mathrm{C} / \mathrm{m}\). A perfectly conducting cylindrical shell, whose axis is the \(z\) axis, surrounds the line charge. The cylinder (of radius \(b\) ), is at ground potential. Under these conditions, the potential function inside the cylinder \((\rhob .(d)\) Find the stored energy in the electric field per unit length in the \(z\) direction within the volume defined by \(\rho>a\), where \(a

Since the conductive cylindrical shell of radius b is at ground potential, its potential is zero. Therefore, we can substitute rho = b into the potential function V(rho) and set it equal to zero to find the constant k: $$ 0 = k - \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln (b) $$ From this we can solve for k: $$ k = \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln (b) $$

The electric field inside the cylindrical shell (rho < b) can be determined from the potential function V(rho) using the relation: $$ \mathbf{E} = -\nabla V $$ In cylindrical coordinates (rho, phi, z), the gradient operator is given by: $$ \nabla = \hat{\rho} \frac{\partial}{\partial \rho} + \hat{\phi} \frac{1}{\rho} \frac{\partial}{\partial \phi} + \hat{z} \frac{\partial}{\partial z} $$ Our potential function V(rho) is independent of phi and z, so we only need to take the partial derivative with respect to rho: $$ \mathbf{E} = \left[ -\hat{\rho} \frac{\partial}{\partial \rho} \left( \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln (\rho) - k \right) \right] $$ Using the above expression for k, we can plug it back into the equation to get the electric field for rho < b: $$ \mathbf{E} = \left[-\hat{\rho} \frac{\partial}{\partial \rho} \left( \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln (\rho) - \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln (b) \right) \right] $$ $$ \mathbf{E} = \left[-\hat{\rho} \frac{\partial}{\partial \rho} \left( \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \ln \left( \frac{\rho}{b} \right) \right) \right] $$ Now, taking the derivative with respect to rho, we get: $$ \mathbf{E} = \left[-\hat{\rho} \frac{\rho_{\ell}}{2 \pi \epsilon_{0}} \frac{1}{\rho} \right] $$ So for rho < b, the electric field is given by: $$ \mathbf{E}(\rho < b) = -\frac{\rho_{\ell}}{2 \pi \epsilon_{0} \rho} \hat{\rho} $$

Since the cylindrical shell is a perfect conductor, the electric field outside the shell (rho > b) is zero: $$ \mathbf{E}(\rho > b) = 0 $$

To find the stored energy in the electric field per unit length, we can use the energy density equation and integrate it over the volume defined by rho > a: $$ \text{Energy per unit length (z)} = \int_{a}^{b} \int_{0}^{2\pi} \frac{1}{2} \epsilon_{0} E^2 \rho \, d\phi d\rho $$ Since we have found the electric field for rho < b, we can plug it back into the energy per unit length equation: $$ \text{Energy per unit length (z)} = \int_{a}^{b} \int_{0}^{2\pi} \frac{1}{2} \epsilon_{0} \left(- \frac{\rho_{\ell}}{2 \pi \epsilon_{0} \rho} \hat{\rho}\right)^2 \rho \, d\phi d\rho $$ Now, we need to integrate with respect to phi and rho: $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{8 \pi \epsilon_{0}} \int_{a}^{b} \int_{0}^{2\pi} \frac{\rho}{\rho^2} \, d\phi d\rho $$ $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{8 \pi \epsilon_{0}} \int_{a}^{b} \int_{0}^{2\pi} \frac{1}{\rho} \, d\phi d\rho $$ $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{8 \pi \epsilon_{0}} \int_{0}^{2\pi} \left[ \ln \left(\frac{\rho}{a} \right)\right]_{a}^{b} \, d\phi $$ $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{8 \pi \epsilon_{0}} \int_{0}^{2\pi} \ln \left(\frac{b}{a} \right) \, d\phi $$ $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{8 \pi \epsilon_{0}} 2\pi \ln \left(\frac{b}{a} \right) $$ Finally, we have: $$ \text{Energy per unit length (z)} = \frac{\rho_{\ell}^2}{4 \epsilon_{0}} \ln \left(\frac{b}{a} \right) $$