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Unformatted text preview: Physics 21 Fall, 2009 Solution to HW-7 24-25 A 5 . 80 F, parallel-plate 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 V d 2 . Inserting the numbers from the problem the calculation is u = 2 400 5 . 00 10- 3 2 = 2 . 83 10- 2 Jm- 3 . 24-38 A parallel-plate 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 ) 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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