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Unformatted text preview: Physics 21 Fall, 2009 Solution to HW7 2425 A 5 . 80 µ F, parallelplate air capacitor has a plate separation of 5 . 00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in units of J / m 3 . The equation for energy density without a dielectric material is given by u = 2 E 2 . The electric field in the region between the parallel plates can by found from the relation V = Ed . Solving for E and combining the equations yields u = 2 µ V d ¶ 2 . Inserting the numbers from the problem the calculation is u = 2 µ 400 5 . 00 × 10 3 ¶ 2 = 2 . 83 × 10 2 Jm 3 . 2438 A parallelplate capacitor has capacitance C = 5 . 00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) what is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates in not to exceed 3 . 00 × 10 4 V/m? (b) A dielectric with K = 2 . 70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the max imum magnitude of charge on each plate if the electric field between the plates is not to exceed 3 . 00 × 10 4 V/m? (a) Without the dielectric, Q = C V and V = Ed . There fore, the charge for the given values of C , E , and d are Q = C Ed = ( 5 . 00 × 10 12 F ) µ 3 . 00 × 10 4 V m ¶ ( . 0015 m) = 2 . 25 × 10 10 C = 225 pC (b) With the dielectric, C increases to...
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 Spring '07
 Hickman
 Charge, Energy, Electric charge, Energy density

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