**Title: Applying Gauss's Law to Determine the Electric Field of a Large Conductor****Objective:** Learn how to use Gauss’s Law to find the electric field at a specified point near a conductor.**Theory and Concept:** Gauss's Law states that the net electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).**Problem Statement:** Determine the electric field at the field point for a semi-infinitely large conductor with a surface charge density \( \sigma \). The distance between the field point and the surface of the conductor is \( d \).**Illustration Description:**- **Diagram Overview:** - **Axes:** - The vertical axis is labeled \( z \). - The horizontal axis is labeled \( x \). - **Conductor:** - A grey rectangle represents a semi-infinitely large conductor along the x-axis. - The top surface of the conductor has positive charges denoted by plus signs (+). - The surface charge density is indicated as \( \sigma \). - **Field Point:** - A point labeled "Field point" is marked above the surface of the conductor. - The field point is at a distance \( d \) from the charged surface along the z-axis.**Solution Approach:**1. **Conceptualize the Problem:** - Consider a Gaussian surface (such as an imaginary box or cylinder) that encloses the charge distribution and extends through the field point.2. **Apply Gauss's Law:** \[ \Phi = E \cdot A = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \] Where: - \( \Phi \) is the electric flux. - \( E \) is the electric field magnitude. - \( A \) is the area through which the field passes. - \( Q_{\text{enclosed}} \) is the charge enclosed. - \( \varepsilon_0 \) is the permittivity of free space.3. **Calculate the Electric Field:** - Use symmetry and the geometry of the problem to simplify calculations. - For a planar surface with charge density \( \sigma \), the electric field \( E \) is: \[ E = \frac{\sigma}{2\varepsilon_

Let d denotes the distance of the field point, E denotes the electric field at that point, σ denotes…

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Use Gauss's law to find the electric field at the field point in the following case. The distance between the field point and the surface of the conductor is d. A semi-infinitely large conductor with surface charge density o. Field point d Conductor