The electric field due to an infinite conducting sheet is given by E = σ / ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space.
Here's a breakdown of the formula and its components:
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E: Represents the electric field strength (measured in Newtons per Coulomb, N/C). It is a vector quantity, indicating both magnitude and direction. The electric field points away from the sheet if the surface charge density is positive and towards the sheet if it is negative.
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σ: Represents the surface charge density (measured in Coulombs per square meter, C/m²). It signifies the amount of electric charge distributed per unit area on the conducting sheet. A higher surface charge density results in a stronger electric field.
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ε₀: Represents the permittivity of free space (approximately 8.854 × 10⁻¹² C²/N·m²). It is a fundamental physical constant that describes the ability of a vacuum to permit electric fields.
Key Points:
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The electric field is uniform and perpendicular to the sheet. This means the magnitude of the electric field is the same at all points in space near the sheet (excluding the edges, which are assumed to be infinitely far away).
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The electric field is independent of the distance from the sheet. Unlike the electric field due to a point charge, which decreases with distance, the electric field from an infinite sheet remains constant.
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For a conducting sheet, the charge resides only on the surface. This is why we consider the surface charge density.
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The electric field is present on both sides of the sheet.
Derivation (Simplified):
While a rigorous derivation involves Gauss's Law, we can understand the basis conceptually. Imagine a Gaussian surface (a closed surface) shaped like a pillbox, with one face on each side of the conducting sheet. Applying Gauss's law (∮ E ⋅ dA = Qenc / ε₀), we find that the electric flux through the sides of the pillbox is zero (since the electric field is parallel to the sides). The electric flux through the top and bottom faces is equal to EA, where A is the area of the face. The enclosed charge, Qenc, is equal to σA. Therefore, 2EA = σA / ε₀, and solving for E gives E = σ / (2ε₀). However, for a conducting sheet, charges redistribute themselves to create an electric field only on one side (effectively doubling the charge density relevant to a single side) such that the electric field inside the conductor is zero. The electric field outside the conductor, on each side is E = σ/ε₀.
Example:
Suppose an infinite conducting sheet has a surface charge density of 2 × 10⁻⁶ C/m². The magnitude of the electric field near the sheet would be:
E = (2 × 10⁻⁶ C/m²) / (8.854 × 10⁻¹² C²/N·m²) ≈ 2.26 × 10⁵ N/C
This electric field would point away from the sheet if the charge density is positive.
