L10) Consider two parallel plate capacitors each with area 1 cm2 and separation 1 mm. These capacitors are placed in series under a power so

Question

L10) Consider two parallel plate capacitors each with area 1 cm2 and separation 1 mm. These capacitors are placed in series under a power source giving 120 V. What is the energy stored on each capacitor?

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Vân Khánh 5 years 2021-09-04T12:05:44+00:00 1 Answers 19 views 0

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  1. lethu
    0
    2021-09-04T12:07:37+00:00
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    Explanation:

    Formula for the capacitance of a parallel plate capacitor is as follows.

               C = \epsilon_{o} \frac{A}{d}

    where,   A = area = 1 cm^{2} = 1 \times 10^{-4} m^{2}

                  d = diameter = 1 mm = 10^{-3} m

           \epsilon_{o} = 8.854 \times 10^{-12} f/m

    Hence, putting the given values into the above formula as follows.

                  C = \epsilon_{o} \frac{A}{d}

                      = 8.854 \times 10^{-12} f/m \times \frac{10^{-4}}{10^{-3}}

                      = 8.854 \times 10^{-11} F

    Now, we will calculate the energy stored in the capacitor as follows.

           U = \frac{1}{2}CV^{2}

               = \frac{1}{2} \times 8.854 \times 10^{-11} \times (120)^{2}

               = 63748.8 \times 10^{-11} J

    or,        = 63.748 \times 10^{-8} J

    Thus, we can conclude that energy stored on each capacitor is 63.748 \times 10^{-8} J.

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