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12 November, 21:09

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.04 cm). What is the energy stored in this new capacitor?

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  1. 13 November, 01:03
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    U_disconnected = 4.4383 * 10^-8 J

    Explanation:

    - The complete information is missing, it is completed below.

    Given:

    - Two parallel plates with Area A = 3481 cm^2

    - Connected to terminals of battery with Vb = 6 V

    - The plates are separated by d = 0.52 cm

    Find:

    The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.04 cm).

    What is the energy stored in this new capacitor?

    Solution:

    - Calculate the total charge Q stored between the plates when the terminals were connected. The charge Q is given by:

    Q = Vb*A*εo / d

    Where, εo = 8.8542 * 10^-12 C / m-V ... Permittivity of Free Space.

    Q = 6*3481*8.8542 * 10^-12 / 0.52*100

    Q = 7.3972*10^-9 C

    - The total energy stored U in the capacitor is given by:

    U = Q*Vb / 2

    U = [ 7.3972*10^-9 ] * 6 / 2

    U = 2.21915*10^-8 J

    - Since the battery is first disconnected, it can do no work and so the charge on the plates remains constant. However moving the plate apart changes the voltage.

    - Voltage is proportional to d, So if d doubles then Voltage doubles.

    - Also Voltage and Total energy U is proportional so U also doubles.

    - So, U:

    U_disconnected = 2*U

    U_disconnected = 2 * 2.21915*10^-8

    U_disconnected = 4.4383 * 10^-8 J
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