(a) Find \(\mathbf{E}\) in the plane \(z=0\) that is produced by a uniform line charge, \(\rho_{L}\), extending along the \(z\) axis over the range \(-L

The electric field in cylindrical coordinates due to a small charge element (dq) at the origin can be expressed as: \(d\textbf{E}= \frac{1}{4 \pi \epsilon_0} \frac{dq}{(ρ^2+z^2)^{3/2}}(\textbf{ρa_ρ}+z\textbf{a_z})\), where ε0 is the permittivity of free space, ρ and z are the cylindrical coordinates, and aρ and az are unit vectors in the cylindrical coordinate system.

The charge element (dq) can be calculated as a product of the linear charge density (ρL) and a charge segment (dz) along the z-axis: \(dq = \rho_L dz\)

Substituting dq into the expression for the electric field, we get: \(d\textbf{E}= \frac{1}{4 \pi \epsilon_0} \frac{\rho_L dz}{(ρ^2+z^2)^{3/2}}(\textbf{ρa_ρ}+z\textbf{a_z})\) Now, we need to integrate this expression to get the total electric field at z=0: \(\textbf{E}= \frac{\rho_L}{4 \pi \epsilon_0} \int_{-L}^{L} \frac{dz}{(ρ^2+z^2)^{3/2}}(\textbf{ρa_ρ}+z\textbf{a_z})\)

To find Eρ and Ez at z=0, we need to separate the integral into two parts: \(E_\rho = \frac{\rho_L}{4 \pi \epsilon_0} \int_{-L}^{L} \frac{\rho dz}{(ρ^2+z^2)^{3/2}}\) and \(E_z = \frac{\rho_L}{4 \pi \epsilon_0} \int_{-L}^{L} \frac{z dz}{(ρ^2+z^2)^{3/2}}\) The integral for Ez=0 at z=0 plane due to symmetry. Now, let's compute the integral for Eρ: \(E_\rho = \frac{\rho_L}{4 \pi \epsilon_0} \int_{-L}^{L} \frac{\rho dz}{(ρ^2+z^2)^{3/2}}\) This integral can be evaluated using the substitution: \(z = ρ \tan{\theta}\) and \(dz = ρ \sec^2{\theta} d\theta\).

With the substitution, the integral becomes: \(E_\rho = \frac{\rho_L}{4 \pi \epsilon_0} \int_{-\arctan{\frac{L}{ρ}}}^{\arctan{\frac{L}{ρ}}} \sec^3{\theta} d\theta\) \(E_\rho = \frac{\rho_L}{2 \pi \epsilon_0} \left[ \frac{1}{2} \sec{\theta} \tan{\theta} - \frac{1}{2} \ln{|\sec{\theta}+\tan{\theta}|} \right]_{-\arctan{\frac{L}{ρ}}}^{\arctan{\frac{L}{ρ}}}\) Now, calculate Eρ for ρ=0.5 L and ρ=0.1 L: \(E_\rho(0.5 L)= \frac{\rho_L}{\pi \epsilon_0 L} \left[ \frac{1}{2} \sec{\theta} \tan{\theta} - \frac{1}{2} \ln{|\sec{\theta}+\tan{\theta}|} \right]_{-\arctan{2}}^{\arctan{2}}\) \(E_\rho(0.1 L)= \frac{\rho_L}{\pi \epsilon_0 L} \left[ \frac{1}{2} \sec{\theta} \tan{\theta} - \frac{1}{2} \ln{|\sec{\theta}+\tan{\theta}|} \right]_{-\arctan{10}}^{\arctan{10}}\)

The electric field due to an infinite line charge is given by: \(E_{\rho(\infty)} = \frac{\rho_L}{2 \pi \epsilon_0 \rho}\) Calculate Eρ for an infinite line charge when ρ=0.5 L and ρ=0.1 L: \(E_{\rho(\infty)}(0.5 L)= \frac{\rho_L}{\pi \epsilon_0 L}\) \(E_{\rho(\infty)}(0.1 L)= \frac{5 \rho_L}{\pi \epsilon_0 L}\)

The percentage error can be calculated as: Percent error = \(\frac{|E_{\rho(\infty)} - E_\rho|}{E_\rho}× 100\) For ρ=0.5 L: Percent error = \(\frac{|\frac{\rho_L}{\pi \epsilon_0 L} - E_\rho(0.5 L)|}{E_\rho(0.5 L)}× 100\) For ρ=0.1 L: Percent error = \(\frac{|\frac{5 \rho_L}{\pi \epsilon_0 L} - E_\rho(0.1 L)|}{E_\rho(0.1 L)}× 100\) Calculate the percentage errors for both cases.