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Math Help - Parallel plate capacitor?

  1. #1
    Super Member fardeen_gen's Avatar
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    Parallel plate capacitor?

    A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected, and the plates pulled apart until their separation is 2d. Derive expression in terms of A, d, and V for:

    (a) the new potential difference
    (b) the initial and final stored energies, U_i and U_f
    (c) the work required to separate the plates

    Answers:
    Spoiler:

    \mbox{(a)}2V
    \mbox{(b)}\frac{\epsilon_0 AV^2}{2d}
    \mbox{(c)}\frac{\epsilon_0 AV^2}{2d}\ \frac{\epsilon_0 AV^2}{2d}\ V = L[\sqrt{y^2 + L^2 - y}]\ E_g = \frac{1}{4\pi \epsilon_0}\left[1 - \frac{y}{\sqrt{y^2 + L^2}}\right]


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  2. #2
    Super Member malaygoel's Avatar
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    The charge on the capacitor will remain constant.

    and Capacitance will vary as <br />
 \frac{\epsilon_0 A}{d}<br />

    hence capacitance will decrease by a factoor of 2.

    Q=CV

    hence, Potential between the plates will increase by a factor of 2.
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