A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacit

Question

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3800.0 μF and is charged to a potential difference of 78.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.84 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 4 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

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Philomena 5 years 2021-08-19T23:19:38+00:00 1 Answers 23 views 0

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  1. yenoanh
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    2021-08-19T23:21:10+00:00
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    Answer:

    – E = 11.55J

    – Q = 0.17C

    – E’ = (1/4)E

    Explanation:

    – To calculate the amount of energy stored in the capacitor, you use the following formula:

    E=\frac{1}{2}CV^2

    C: capacitance = 3800.0*10^-6F

    V: potential difference = 78.0V

    E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J

    The energy stored in the capacitor is 11.55J

    – If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

    E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C

    The charge on the capacitor is 0.17C

    – If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

    E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\

    A: area of the plates

    d: distance between plates

    If the distance between plates is increased by a factor of 4, you have:

    E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E

    Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

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