Given the surface charge density, \(\rho_{s}=2 \mu \mathrm{C} / \mathrm{m}^{2}\), existing in the region \(\rho<\) \(0.2 \mathrm{~m}, z=0\), find \(\mathbf{E}\) at \((a) P_{A}(\rho=0, z=0.5) ;(b) P_{B}(\rho=0, z=-0.5)\). Show that \((c)\) the field along the \(z\) axis reduces to that of an infinite sheet charge at small values of \(z ;(d)\) the \(z\) axis field reduces to that of a point charge at large values of \(z\).

First, we need to find the electric field using Gauss's Law. Gauss's Law states that the flux through any closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space. Mathematically, it can be written as: \(\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{encl}}{\epsilon_0}\) In order to find the electric field, we must integrate over a cylindrical Gaussian surface so that we can take advantage of the symmetry of the problem. The differential flux through the side surface of the cylinder is given by \(d\Phi = E \cdot dA\) where \(dA = 2\pi \rho d\rho dz\). Thus, the total flux through the cylinder is: \(\Phi = \int\int \mathbf{E} \cdot d\mathbf{A} = \int\int E \cdot 2\pi \rho d\rho dz\) Now, we must find \(Q_{encl}\). As we know the surface charge density (\(\rho_s\)) and the limits of the region where it exists, we can compute \(Q_{encl}\) as follows: \(Q_{encl} = \rho_s \int\int dA = \rho_s \int\int 2\pi \rho d\rho dz = 2\pi \rho_s \int_0^{0.2} \rho d\rho \int_0^{2\pi} dz\) By equating the total flux from Gauss's Law to \(Q_{encl}/\epsilon_0\), we can solve for the electric field at points A and B, inside and outside the region where the surface charge exists.

To see that the field along the z-axis reduces to that of an infinite sheet charge at small values of z, we must compare the obtained electric field to that of an infinite sheet: \(E_{sheet} = \frac{\rho_s}{2\epsilon_0}\) If the obtained electric field reduces to this value for small values of z, then it behaves like an infinite sheet charge.

To demonstrate that the obtained electric field reduces to that of a point charge for large values of z, we compare the obtained field to the field of a point charge: \(E_{point} = \frac{Q_{encl}}{4\pi\epsilon_0z^2}\) If our electric field exhibits the same dependence for large values of z, then it behaves like a point charge in that scenario. Now, use the obtained expressions for the electric field at points A and B to compare the fields, and verify the conditions (c) and (d).