Each plate of an ideal air-filled parallel-plate capacitor has an area of 0.0020 $m ^ { 2 }$ , and the separation of the plates is $0.090 \mathrm {~mm}$ An electric field of $2.1 \times 10^6 \mathrm {~N} / \mathrm { C }$ is present between the plates. What is the surface charge density on the plates? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

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Quizplus · Accepted Answer

The surface charge density is given by:$$\sigma = \epsilon_0 E$$where:$$\sigma$$ is the surface charge density$$\epsilon_0$$ is the permittivity of free space$$E$$ is the electric fieldSubstituting the given values into the equation, we get:$$\sigma = (8.85 imes 10^{-12} \mathrm{ C}^2/\mathrm{N} \cdot \mathrm{m}^2)(2.1 imes 10^6 \mathrm{ N} / \mathrm{C}) = 19 \mu \mathrm{C}/\mathrm{m}^2$$Therefore, the surface charge density on the plates is 19 µC/m².

Each Plate of an Ideal Air-Filled Parallel-Plate Capacitor Has an Area