The electric field due to an infinite sheet of charge is uniform, perpendicular to the sheet, and has a magnitude of σ / (2ε₀), where σ is the surface charge density and ε₀ is the permittivity of free space.
Explanation
The electric field produced by an infinite sheet of charge is a classic problem in electromagnetism, demonstrating the application of Gauss's Law. Here's a breakdown:
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Uniformity: The electric field strength is the same at all points equidistant from the sheet. This is because the sheet is infinite, so there are no edge effects.
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Direction: The electric field lines are perpendicular to the sheet. If the charge density is positive, the electric field points away from the sheet. If the charge density is negative, the electric field points toward the sheet.
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Magnitude: The magnitude of the electric field (E) can be derived using Gauss's Law:
E = σ / (2ε₀)
Where:
- σ (sigma) is the surface charge density (charge per unit area) of the sheet (Coulombs per square meter, C/m²).
- ε₀ (epsilon naught) is the permittivity of free space (approximately 8.854 x 10⁻¹² C²/N·m²).
Derivation using Gauss's Law
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Choose a Gaussian Surface: Imagine a cylindrical Gaussian surface that is perpendicular to the charged sheet. The cylinder extends through the sheet, with its ends equidistant from the sheet.
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Apply Gauss's Law: Gauss's Law states that the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space:
∮ E · dA = Q_enclosed / ε₀
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Calculate the Electric Flux: Since the electric field is perpendicular to the sheet, the electric flux only passes through the two ends of the cylinder. Let the area of each end be A. The electric flux through each end is E * A. Therefore, the total electric flux is 2EA.
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Calculate the Enclosed Charge: The charge enclosed by the Gaussian surface is the surface charge density (σ) multiplied by the area of the sheet enclosed by the cylinder's end (A). So, Q_enclosed = σA.
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Solve for the Electric Field: Substituting the expressions for electric flux and enclosed charge into Gauss's Law:
2EA = σA / ε₀
Solving for E:
E = σ / (2ε₀)
Key Points
- The electric field is independent of the distance from the sheet. This is a characteristic of an infinite sheet. Real-world sheets of charge will have edge effects, so this approximation is most accurate close to the sheet.
- The electric field is uniform on both sides of the sheet, pointing away from the sheet if the charge is positive, and toward the sheet if the charge is negative.
