Example : A square conducting plate of side length \( 2 a \), positioned in the \( x-y \) plane and centred at the origin, is charged with a uniform surface charge density \( \rho_{0} \). Determine the electric field intensity at \( (0,0, z) \) due to this surface charge distribution. [Given \( \int_{0}^{a} \int_{0}^{a} \frac{d x d y}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}=\frac{1}{z} \tan ^{-1} \frac{a^{2}}{\left.z \sqrt{2 a^{2}+z^{2}}\right]} \) Due to the symmetry of the charge distribution about the \( z \)-axis, the \( x \) and \( \begin{array}{l}\text { and } \\ \text { components vanish } \\ \text { and only,one quarter of } \\ \text { the plate can be }\end{array} \) considered.

To find the electric field intensity at the point \( (0,0,z) \), we start by considering the contribution from the surface charge on a small element of the plate. Due to symmetry, we can take one quarter of the square plate and multiply the result by four. The vertical component of the electric field from this small area can be derived from Coulomb’s law, integrating over the charged area while using the provided integral for the electric field contribution. Once we derive the expression, we must remember that the total electric field will combine all contributions. The final results will yield an electric field intensity that diminishes with distance \( z \) as you move away from the plate, typically behaving like \( E \propto \frac{1}{z^2} \) for points far from the charged plate, illustrating the point charge behavior as distance increases!