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The electric field between the plates of a paper-separated capacitor is 8.18×104 . The plates are 2.20 apart, and the charge on each plate is 0.795 .Determine the capacitance of this capacitor. Determine the area of each plate.

  • Physics -

    E= 8.18•10⁴ V/m
    d=2.2 mm
    q=0.775 μ C
    ε (paper)= 2.3
    (a) C=q/U

    C=q/Ed = ...
    ε₀=8.85•10⁻¹² F/m

    C= ε₀εA/d =>

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