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Unformatted text preview: # 13.5 mJ 2 2 (b) BU # 13.5 mJ ! 3.6 mJ # 9.9 mJ . The energy increased. EVALUATE:! The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted. U # 1 QV , so the stored energy increases. 2 IDENTIFY:! Gauss’s law in dielectrics has the same form as in vacuum except that the electric field is multiplied by a factor of K and the charge enclosed by the Gaussian surface is the free charge. The capacitance of an object depends on its geometry. (a) SET UP:! The capacitance of a parallel-plate capacitor is C # K !0 A/d and the charge on its plates is Q = CV. Capacitance and Dielectrics 24-13 EXECUTE:! First find the capacitance: C# K !0 A (2.1)(8.85 " 10!12 C2 /N + m2 )(0.0225 m2 ) # # 4.18 " 10!10 F . d 1.00 " 10!3 m Now find the charge on the plates: Q # CV # (4.18 " 10!10 F)(12.0 V) = 5.02 " 10!9 C . (b) SET UP:! Gauss’s law within the dielectric gives KEA # Qfree /!0 . EXECUTE:! Solving for E gives Q 5.02 " 10!9 C E # free # # 1.20 " 104 N/C KA!0 (2.1)(0.0225 m 2 )(8.85 " 10!12 C 2 /N + m 2 ) (c) SET UP:! Without the Teflon and the voltage source, the charge is unchanged but the potential increases, so C # !0 A/d and Gauss’s law now gives EA # Q/!0 . EXECUTE:! First find the capacitance: !0 A (8.85 " 10!12 C2 /N + m 2 )(0.0225 m2 ) # # 1.99 " 10!10 F. d 1.00 " 10!3 m C# The potential difference is V # E# 24.49. Q 5.02 " 10!9 C # # 25.2 V. From Gauss’s law, the electric field is C 1.99 " 10!10 F Q 5.02 " 10!9 C # # 2.52 " 10 4 N/C. !12 !0 A (8.85 " 10 C 2 /N + m 2 )(0.0225 m 2 ) EVALUATE:! The dielectric reduces the electric field inside the capacitor because the electric field due to the dipoles of the dielectric is opposite to the external field due to the free charge on the plates. IDENTIFY:! Apply Eq.(24.23) to calculate E. V = Ed and C = Q/V apply whether there is a dielectric between the plates or not. (a)...
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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