The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between

Question

The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?

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Thu Thủy 3 years 2021-09-04T17:40:26+00:00 1 Answers 2 views 0

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  1. Philomena
    0
    2021-09-04T17:41:33+00:00
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    Explanation:

    The given data is as follows.

            Dielectric constant, K = 3.0

       Area of the plates (A) = 0.021 m^{2}

       Distance between plates (d) = 2.75 \times 10^{-3} m

      Maximum electric field (E) = 3.25 \times 10^{5} V/m

    Now, we will calculate the capacitance as follows.

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

                     = \frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}

                     = \frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}

                     = 0.203 \times 10^{-9} F

    Formula to calculate electric charge is as follows.

                   E = \frac{\sigma}{k \epsilon_{o}}

    or,           Q = E \times k \times \epsilon_{o}A      (as \frac{\sigma}{\epsilon_{o}} = \frac{Q}{A})

                       = 3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021

                       = 181.2 \times 10^{-9} C

    Formula to calculate the energy is as follows.

                      U = \frac{1 \times Q^{2}}{2 \times C}

                         = \frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}

                         = \frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}

                         = 9835.67 \times 10^{-9}

    or,                  = 98.35 \times 10^{-7} J

    Thus, we can conclude that the maximum energy that can be stored in the capacitor is 98.35 \times 10^{-7} J.

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