The parallel plates in a capacitor, with a plate area of 9.50 cm2 and an air-filled separation of 3.40 mm, are charged by a 7.60 V battery.

Question

The parallel plates in a capacitor, with a plate area of 9.50 cm2 and an air-filled separation of 3.40 mm, are charged by a 7.60 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 9.50 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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Bình An 5 years 2021-07-28T18:50:15+00:00 2 Answers 18 views 0

Answers ( )

  1. vankhanh
    0
    2021-07-28T18:51:19+00:00
    Reply

    Answer:

    Explanation:

    capacitance of capacitor = ε₀ A / d

    ε₀ = 8.85 x 10⁻¹² , A is area of plate and d is plate separation .

    = 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 3.4 x 10⁻³

    C  = 24.73 x 10⁻¹³

    capacitance of capacitor after increase in plate separation

    = 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 9.5  x 10⁻³

    = 8.85 x 10⁻¹³

    initial charge = capacitance x potential

    = 24.73 x 10⁻¹³ x 7.6 C

    potential difference after increased separation

    = initial charge / increased capacitance

    = 24.73 x 10⁻¹³ x 7.6 / (8.85 x 10⁻¹³)

    = 21.23 V .

    b ) initial stored energy

    = 1/2 C V²

    = .5 x 24.73 x 10⁻¹³ x 7.6²

    = 714.2 x 10⁻¹³ J

    c ) final stored energy

    1/2 C V²

    = .5 x 8.85  x 10⁻¹³ x 21.23 ²

    = 1994.4 x 10⁻¹³ J

    d ) work done in separation of plate

    =2 x  increase in stored energy

    =  2 x (1994.4 – 714.2 ) x 10⁻¹³ J

    = 2560.4 x 10⁻¹³ J .

  2. Helga
    0
    2021-07-28T18:51:33+00:00
    Reply

    Answer:

    Explanation:

    Area of plates, A = 9.50 cm²

    separation, d = 3.40 mm

    Voltage, V = 7.60 V

    new separation, d’ = 9.5 mm

    The formula of the parallel plate capacitance is given by

    C = \frac{\epsilon_{0}A}{d}

    C = \frac{8.854\times 10^{-12}\times 9.5\times 10^{-4}}{3.4\times 10^{-3}}

    C = 2.44 x 10^-12 F

    initial charge, q = C x V

    q = 2.44 x 10^-12 x 7.6 = 1.88 x 10^-11 C

    (a)

    As the battery is disconnected so the charge remains same..

    The new capacitance, C’ is given by

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

    C' = \frac{8.854\times 10^{-12}\times 9.5\times 10^{-4}}{9.5\times 10^{-3}}

    C’ = 8.854 x 10^-13 F

    Let the new potential difference is V’

    V’ = q / C’

    V’ = (1.88 x 10^-11) / (8.854 x 10^-13) = 21.23 V

    (b) Initial energy

    U = 0.5 x CV²

    U = 0.5 x 2.44 x 10^-12 x 7.6 x 7.6 = 7.05 x 10^-11 J

    (c) Final energy

    U’ = 0.5 x C’ x V’²

    U’ = 0.5 x  8.854 x 10^-13 x 21.23 x 21.23 = 1.995 x 10^-10 J

    (d) Work done = change in potential energy

    W = U’ – U

    W = 1.995 x 10^-10 – 7.05 x 10^-11

    W = 12.9 x 10^-11 Joule

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