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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-7 24-25 A 6 . 60 F, parallel-plate, air capacitor has a plate separation of 3 . 00 mm and is charged to a potential difference of 300 V. Calculate the energy density in the region between the plates. Energy density is calculated using the formula u = 1 2 E 2 . In this problem, we know the voltage on the capacitor, not the electric field strength between the plates. But the volt- age and electric field strength are related. Because the elec- tric field between the plates is uniform and directed straight across the gap between the plates, it is easy to show that V = Ed where d is the separation between the plates. Do- ing the substitutions and evaluating for the given values, we get u = 1 2 parenleftbigg V d parenrightbigg 2 = 1 2 parenleftbigg 8 . 85 10- 12 C 2 N m 2 parenrightbiggparenleftbigg 300 V . 003 m parenrightbigg 2 = 4 . 43 10- 2 J / m 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 is 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 ) parenleftbigg 3 . 00 10 4 V m parenrightbigg ( . 0015 m) = 2 . 25 10- 10 C = 225 pC (b) With the dielectric, C increases to KC , and V = Ed still holds. The charge is K times the previous value: Q = KC Ed = 2 . 7 ( 2 . 25 10- 10 C )...
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