Find the energy W of the electrostatic field of the lay

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Find the energy W of the electrostatic field of the capacitor laminate flat, square plates which S = 400cm ^ 2, the thickness of the first hard rubber layer capacitor d1 = 0,02 cm, the second layer of glass d2 = 0,07 cm. The dielectric constant e = ebony 3, E-glass = 7. The charge of the capacitor is Q = 1,0 * 10 ^ -8 Kl.

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Task 31279. Detailed solution shorthand terms, formulas, laws used in the decision, the withdrawal of the calculation formula and the response.
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