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Unformatted text preview: Physics 21 Fall, 2008 Solution to HW-7 24-27 A capacitor with a capacitance of 460 µ F is charged to a voltage of 270 V. Then a wire is connected between the plates. How many joules of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire? C = 470 µ F; V = 270 V; U cap = 1 2 CV 2 = 33 . 5 J 24-39 Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E = 3 . 00 × 10 5 V/m. When the space is filled with dielectric, the electric field is E K = 2 . 10 × 10 5 V/m. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant? (a) The magnitude of the surface charge density is related to the field inside a capacitor by E = σ/ . The charge density on the plates when the dielectric is absent is σ = E . In order for the field to diminish to E K , there must be an opposite surface charge σ dielectric on the dielectric to reduce the net surface charge near the capacitor plates to E K . That is, E K = ( σ − σ dielectric ) / = E − σ dielectric / Solving for σ dielectric , we find σ dielectric = ( E − E K ) = 7 . 96 × 10- 7 c / m 2 ....
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