The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.78 cm). What is the energy stored in this

Question

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.78 cm). What is the energy stored in this new capacitor?

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Vodka 3 months 2021-09-02T05:12:52+00:00 1 Answers 1 views 0

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    2021-09-02T05:14:23+00:00

    Answer:

    The energy stored in this new capacitor is 4.4514\times10^{-9}\ J

    Explanation:

    Suppose, Two parallel plates, each having area A = 2180 cm² are connected to the terminals of a battery of voltage V_{b}= 6\ V as shown. The plates are separated by a distance d = 0.39 cm.

    We need to calculate the charge

    Using formula of capacitance

    C=\dfrac{Q}{V}

    \dfrac{Q}{V}=\dfrac{\epsilon_{0}A}{d}

    Q=V\times\dfrac{\epsilon_{0}A}{d}

    Put the value into the formula

    Q=6\times\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.39\times10^{-2}}

    Q=2.968\times10^{-9}\ C

    The distance between the plates is doubled.

    We need to calculate the new capacitance

    Using formula of capacitance

    C'=\dfrac{\epsilon_{0}A}{d}

    Put the value into the formula

    C'=\dfrac{8.85\times10^{-12}\times2180\times10^{-4}}{0.78\times10^{-2}}

    C'=2.473\times10^{-10}\ F

    We need to calculate the energy stored in this new capacitor

    Using formula of energy

    U=\dfrac{1}{2}C'V^2

    Put the value into the formula

    U=\dfrac{1}{2}\times2.473\times10^{-10}\times(6)^2

    U=4.4514\times10^{-9}\ J

    Hence, The energy stored in this new capacitor is 4.4514\times10^{-9}\ J

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