An air filled parallel plate capacitor has plates of area 0.0406 m^2 and seperation of 2.5*10^-6 m.
a)What is the capacitence for the setup?
b)Our capacitor is connected to a 550 v battery, how much energy is stored in the capacitor?
c) If -1 mJ of work is performed by the capapcitor's plates but it is never disconnected from the 550 V battery, what is the new seperation of the plates?

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  1. a) C = epsilon*A/d
    Epsilon = 8.85*10^-12 is the permittivity of free space.
    I get C = 1.44*10^-7 F

    b) Energy = (1/2) C V^2
    I get 2.17*10^-2 J

    c) If - 1 mJ is done BY the plates, then +1 mJ (0.001 J) is done ON the plates, by whatever force is changing the plate spacing. Electrical energy may also flow between capacitor and the battery while the plates are moved.

    To get the work required, multiply the force between the plates by x = 0.001 m. The force is EQ*x were
    E = V/d is the electric field and Q = CV is the charge.

    Work = 0.001 J = (CV^2/d) * x

    Solve for x and add it to the origial plate spacing. The plate spacing will be increased, but the stored energy in the capacitor will be less, because C will be reduced at the same voltage.

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