Task 31260.
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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 constants e1 = 3 ebony, glass e2 = 7. The charge of the capacitor is Q = 1,0 * 10 ^ -8 Kl.
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- Content description 15,69 kB
- Updated on the site 17.05.2015
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Detailed solution shorthand terms, formulas, laws used in the decision, the withdrawal of the calculation formula and the response.
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