Two large parallel plates are separated by 1.0 mm of air.If the potential difference between them is 3.0 V,what is the magnitude of their surface charge densities? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

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Quizplus · Accepted Answer

The electric field (E) between the plates is V/d = 3.0 V / 0.001 m = 3000 V/m. The surface charge density (σ) is given by σ = ε₀E = 8.85 × 10⁻¹² C²/N∙m² × 3000 V/m = 2.655 × 10⁻⁸ C/m², which rounds to 2.7 × 10⁻⁸ C/m².

Two Large Parallel Plates Are Separated by 1