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 determine the electric field intensity \(\mathbf{E}\) at the point \((0, 0, z)\) due to a uniformly charged square conducting plate, follow the detailed steps below. ### **Given:** - **Plate Specifications:** - Shape: Square - Side Length: \(2a\) - Position: \(x\text{-}y\) plane, centered at the origin - Surface Charge Density: \(\rho_0\) - **Point of Interest:** \((0, 0, z)\) - **Integral Provided:** \[ \int_{0}^{a} \int_{0}^{a} \frac{dx\, dy}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}} = \frac{1}{z} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \] ### **Objective:** Determine the electric field intensity \(\mathbf{E}\) at the point \((0, 0, z)\) due to the surface charge distribution on the square plate. ### **Solution:** 1. **Symmetry Considerations:** - Due to the square's symmetry about the \(z\)-axis, the \(x\) and \(y\) components of the electric field cancel out. Hence, only the \(z\)-component (\(E_z\)) remains. - To simplify the calculation, consider only one quarter of the plate (e.g., \(x \geq 0\) and \(y \geq 0\)) and multiply the result by 4. 2. **Expressing the Electric Field:** The electric field \(E_z\) at point \((0, 0, z)\) is given by: \[ E_z = \frac{\rho_0}{4\pi \epsilon_0} \times 4 \int_{0}^{a} \int_{0}^{a} \frac{z\, dx\, dy}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}} \] Simplifying: \[ E_z = \frac{\rho_0}{\pi \epsilon_0} \int_{0}^{a} \int_{0}^{a} \frac{z\, dx\, dy}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}} \] 3. **Evaluating the Double Integral:** Using the provided integral: \[ \int_{0}^{a} \int_{0}^{a} \frac{dx\, dy}{\left(x^{2} + y^{2} + z^{2}\right)^{\frac{3}{2}}} = \frac{1}{z} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \] Substitute this into the expression for \(E_z\): \[ E_z = \frac{\rho_0}{\pi \epsilon_0} \times z \times \frac{1}{z} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \] Simplifying: \[ E_z = \frac{\rho_0}{\pi \epsilon_0} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \] 4. **Final Expression for the Electric Field:** \[ \mathbf{E} = \frac{\rho_0}{\pi \epsilon_0} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \hat{\mathbf{z}} \] ### **Conclusion:** The electric field intensity at the point \((0, 0, z)\) due to the uniformly charged square conducting plate is directed along the \(z\)-axis and is given by: \[ \mathbf{E} = \frac{\rho_0}{\pi \epsilon_0} \tan^{-1}\left(\frac{a^{2}}{z \sqrt{2a^{2} + z^{2}}}\right) \hat{\mathbf{z}} \] This expression accounts for the contribution of the entire plate by leveraging symmetry and the provided integral, ensuring an accurate calculation of the electric field at the specified point.