An air-filled capacitor consists of two parallel plates, each with an area of 8 cm^2 , separated by a distance 2.9 mm. A 22 V potential diff

Question

An air-filled capacitor consists of two parallel plates, each with an area of 8 cm^2 , separated by a distance 2.9 mm. A 22 V potential difference is applied to these plates.
What is the magnitude of the surface charge density on each plate?

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Kiệt Gia 6 months 2021-08-04T07:11:48+00:00 1 Answers 2 views 0

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    2021-08-04T07:13:22+00:00

    Answer:

    The magnitude of the surface charge density on each plate is 6.714 x 10⁻ C/m²

    Explanation:

    Given;

    area of the parallel plates, A = 8 cm² = 8 x 10⁻⁴ m²

    distance between the plates, d = 2.9 mm = 0.0029 m

    potential difference applied to the plates, V = 22 V

    The electric field between the plates is given by;

    E = \frac{V}{d} \\\\E = \frac{22}{2.9*10^{-3}}\\\\E = 7586.207 \ V/m

    The surface charge density is given by;

    σ = ε₀E

    σ = (8.85 x 10⁻¹²)(7586.207)

    σ = 6.714 x 10⁻ C/m²

    Therefore, the magnitude of the surface charge density on each plate is 6.714 x 10⁻ C/m²

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