Calculating Electric Field From Square Charge Sheet A Detailed Solution

In electromagnetism, understanding the electric field generated by charge distributions is crucial. This article delves into a specific problem: calculating the electric field E at the origin due to a square sheet of charge lying in the z = -3 m plane. The sheet is defined by -2 ≤ x ≤ 2 m and -2 ≤ y ≤ 2 m, with a charge density ρ_s = 2(x² + y² + 9)^(3/2) nC/m². We will explore the concepts, methodology, and detailed calculations involved in solving this problem. This problem serves as a great example of applying integral calculus and vector analysis in electromagnetics. In the following sections, we will break down the problem step by step, ensuring a clear and comprehensive understanding of the solution.

The electric field E due to a continuous charge distribution is given by the integral form of Coulomb's law. For a surface charge distribution, the electric field at a point is calculated by integrating the contributions from infinitesimal charge elements over the entire surface. The formula for the electric field dE due to an infinitesimal charge element dQ is:

dE = (1 / (4πε₀)) * (dQ / R²) * R̂

Where:

• ε₀ is the permittivity of free space (approximately 8.854 × 10⁻¹² F/m).
• dQ is the infinitesimal charge element.
• R is the distance from the charge element to the point of interest.
• R̂ is the unit vector pointing from the charge element to the point of interest.

For a surface charge density ρ_s, the infinitesimal charge element dQ can be expressed as:

dQ = ρ_s * dA

Where dA is the infinitesimal area element. In our case, since the charge sheet lies in the z = -3 m plane, it's convenient to express dA in Cartesian coordinates as dx dy. The position vector r' from the origin to the infinitesimal charge element is given by r' = xî + yĵ - 3k̂, and the position vector r of the point of interest (the origin) is r = 0. The distance R is the magnitude of the vector difference R = r - r' , and the unit vector R̂ is given by R̂ = R / R. Understanding these fundamental concepts is crucial for setting up the integral to calculate the electric field.

To solve for the electric field at the origin, we need to set up the integral by identifying all the components. Given the charge density ρ_s = 2(x² + y² + 9)^(3/2) nC/m² and the area element dA = dx dy, the infinitesimal charge element dQ is:

dQ = 2(x² + y² + 9)^(3/2) × 10⁻⁹ dx dy

The position vector r' from a point on the sheet to the origin is given by r' = xî + yĵ - 3k̂. The magnitude R of this vector is R = √(x² + y² + 9). The unit vector R̂ pointing from the charge element to the origin is:

R̂ = - (xî + yĵ - 3k̂) / √(x² + y² + 9)

Now, we can express the infinitesimal electric field dE as:

dE = (1 / (4πε₀)) * (2(x² + y² + 9)^(3/2) × 10⁻⁹ dx dy / (x² + y² + 9)) * (- (xî + yĵ - 3k̂) / √(x² + y² + 9))

Simplifying, we get:

dE = - (1 / (2πε₀)) × 10⁻⁹ * (xî + yĵ - 3k̂) dx dy

The total electric field E is the integral of dE over the square sheet defined by -2 ≤ x ≤ 2 m and -2 ≤ y ≤ 2 m. We will integrate each component separately due to the vector nature of the electric field.

The electric field E has three components: Ex, Ey, and Ez. We calculate each component by integrating the corresponding component of dE over the given limits. The total electric field E is the vector sum of these components.

Ex Component

The x-component of the electric field, Ex, is given by:

Ex = ∫∫ dEx = - (1 / (2πε₀)) × 10⁻⁹ ∫[-2 to 2] ∫[-2 to 2] (x dx dy)

The integral of x over a symmetric interval from -2 to 2 is zero. Therefore, Ex = 0. This result is expected due to the symmetry of the charge distribution about the yz-plane. The symmetry argument simplifies the calculation significantly. The physical interpretation is that for every charge element on the right side of the yz-plane, there is a corresponding charge element on the left side that produces an equal and opposite x-component of the electric field, resulting in cancellation.

Ey Component

Similarly, the y-component of the electric field, Ey, is given by:

Ey = ∫∫ dEy = - (1 / (2πε₀)) × 10⁻⁹ ∫[-2 to 2] ∫[-2 to 2] (y dx dy)

Again, the integral of y over a symmetric interval from -2 to 2 is zero. Therefore, Ey = 0. This result is also expected due to the symmetry of the charge distribution about the xz-plane. For every charge element above the xz-plane, there is a corresponding charge element below the xz-plane that produces an equal and opposite y-component of the electric field. This cancellation leads to a net Ey component of zero.

Ez Component

The z-component of the electric field, Ez, is given by:

Ez = ∫∫ dEz = (3 / (2πε₀)) × 10⁻⁹ ∫[-2 to 2] ∫[-2 to 2] dx dy

This integral is straightforward to evaluate:

Ez = (3 × 10⁻⁹ / (2πε₀)) ∫[-2 to 2] [x] [-2 to 2] dy

Ez = (3 × 10⁻⁹ / (2πε₀)) ∫[-2 to 2] (2 - (-2)) dy

Ez = (3 × 10⁻⁹ / (2πε₀)) ∫[-2 to 2] 4 dy

Ez = (12 × 10⁻⁹ / (2πε₀)) [y] [-2 to 2]

Ez = (12 × 10⁻⁹ / (2πε₀)) (2 - (-2))

Ez = (48 × 10⁻⁹) / (2πε₀)

Substituting the value of ε₀ (8.854 × 10⁻¹² F/m), we get:

Ez = (48 × 10⁻⁹) / (2π × 8.854 × 10⁻¹²)

Ez ≈ 863.3 N/C

The total electric field E at the origin is the vector sum of its components:

E = Exî + Eyĵ + Ezk̂

Since Ex = 0 and Ey = 0, the electric field is:

E = 863.3k̂ N/C

The electric field at the origin is solely in the z-direction, pointing away from the charge sheet. This result aligns with the physical intuition that the electric field should be perpendicular to the plane of the charge sheet and directed away from it due to the positive charge density.

In this article, we calculated the electric field at the origin due to a square sheet of charge using integral calculus and vector analysis. The problem was broken down into steps, starting from the theoretical background of Coulomb's law and the electric field due to continuous charge distributions. We set up the integral by identifying the infinitesimal charge element, the distance, and the unit vector. The integral was then evaluated for each component of the electric field, and the symmetry of the charge distribution was used to simplify the calculations. The final result shows that the electric field at the origin is approximately 863.3 N/C in the positive z-direction.

This problem illustrates the application of electromagnetic theory to practical scenarios. Understanding how to calculate electric fields due to charge distributions is essential in many areas of physics and engineering, such as capacitor design, electromagnetic shielding, and the study of charged particle motion. The methodology presented here can be extended to calculate electric fields for more complex charge distributions and geometries. Mastering these techniques provides a solid foundation for further studies in electromagnetics and related fields.