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? (?₀ = 8.85 × 10^-12C²/N ? m²)

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

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