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27 March, 09:44

A parallel-plate air capacitor of area A = 28.1 cm2 and plate separation of d = 3.80 mm is charged by a battery to a voltage of 69.0 V. What is the charge on the capacitor?

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  1. 27 March, 10:00
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    4.52*10⁻¹⁰ C

    Explanation:

    The charge stored in a capacitor is given as,

    Q = CV ... Equation 1

    Where Q = Charge stored in a capacitor, C = Capacitance of the capacitor, V = Voltage.

    But,

    C = e₀A/d ... Equation 2

    Where e₀ = permitivity of free space, A = Area of the plates, d = distance of separation of the plates

    Substitute equation 2 into equation 1

    Q = e₀AV/d ... Equation 3

    Given: A = 28.1 cm² = 0.00281 m², V = 69.0 V, d = 3.8 mm = 0.0038 m

    Constant: e₀ = 8.85*10⁻¹² F/m.

    Substitute into equation 3

    Q = 8.85*10⁻¹²*0.00281*69/0.0038

    Q = 4.52*10⁻¹⁰ C.

    Hence the charge on the capacitor = 4.52*10⁻¹⁰ C
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