The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery.

Question

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates

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Jezebel 1 year 2021-08-21T07:25:04+00:00 1 Answers 61 views 0

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    2021-08-21T07:26:23+00:00

    Answer:

    (a). The charge is 1.5045\times10^{-11}\ C

    (b). The initial stored energy is 4.5135\times10^{-11}\ J

    (c).  The final stored energy is 12.036\times10^{-11}\ J

    (d). The work required to separate the plates is 7.5225\times10^{-11}\ J

    Explanation:

    Given that,

    Area = 8.50 cm²

    Distance = 3.00 mm

    Potential = 6.00 V

    Distance without discharge = 8.00 mm

    (a). We need to calculate the capacitance

    Using formula of capacitance

    C_{1}=\dfrac{\epsilon_{0}A}{d}

    Put the value into the formula

    C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}

    C_{1}=2.5075\times10^{-12}\ F

    We need to calculate the charge

    Using formula of charge

    Q=CV

    Put the value into the formula

    Q=2.5075\times10^{-12}\times6.00

    Q=1.5045\times10^{-11}\ C

    (b). We need to calculate the initial stored energy

    Using formula of initial energy

    E_{i}=\dfrac{1}{2}\times CV^2

    E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36

    E_{i}=4.5135\times10^{-11}\ J

    (c). We need to calculate the capacitance

    Using formula of capacitance

    C_{2}=\dfrac{\epsilon_{0}A}{d}

    Put the value into the formula

    C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}

    C_{2}=9.403\times10^{-13}\ F

    We need to calculate the final stored energy

    Using formula of initial energy

    E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}

    E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}

    E_{f}=12.036\times10^{-11}\ J

    (d). We need to calculate the work done

    Using formula of work done

    W=E_{f}-E_{i}

    Put the value in the formula

    W=12.036\times10^{-11}-4.5135\times10^{-11}

    W=7.5225\times10^{-11}\ J

    Hence, (a). The charge is 1.5045\times10^{-11}\ C

    (b). The initial stored energy is 4.5135\times10^{-11}\ J

    (c).  The final stored energy is 12.036\times10^{-11}\ J

    (d). The work required to separate the plates is 7.5225\times10^{-11}\ J

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