Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2.0 m.
Question
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2.0 m.
Solution
The electric field due to an infinite plane of charge is given by the equation:
E = σ / (2ε₀)
where σ is the surface charge density and ε₀ is the permittivity of free space.
Given that σ = 0.20 nC/m² = 0.20 x 10^-9 C/m² and ε₀ = 8.85 x 10^-12 C²/N·m², we can substitute these values into the equation:
E = (0.20 x 10^-9 C/m²) / (2 x 8.85 x 10^-12 C²/N·m²)
Solving this gives:
E = 11.3 N/C
The direction of the electric field is perpendicular to the plane of charge and points away from the plane if the charge is positive. The magnitude of the electric field does not depend on the distance from the plane, so the electric field at z = 2.0 m is also 11.3 N/C.
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