Homework 6-solutions - soto (rrs946) Homework 6 Hoffmann...

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Unformatted text preview: soto (rrs946) Homework 6 Hoffmann (56605) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The electric potential energy stored in a ca- pacitor is U = 1 2 Q 2 C = 1 2 QV = 1 2 C V 2 . a) An isolated capacitor has a dielectric slab between its plates. b) The capacitor is charged by a battery. c) After the capacitor is fully charged, the battery remains connected. d) Finally, the dielectric slab is moved out of the capacitor. Let U out denote the electric potential en- ergy of the capacitor when the dielectric is out of the capacitor. The work required to remove the dielectric from the capacitor is 1. W = 3 2 ( - 1) U out 2. W = parenleftbigg 3 2- 2 3 parenrightbigg U out 3. W = parenleftbigg 1- 1 parenrightbigg U out 4. W = parenleftbigg 3 2 - 2 3 parenrightbigg U out 5. W = parenleftbigg 1 - 1 parenrightbigg U out 6. W = 3 2 parenleftbigg 1- 1 parenrightbigg U out 7. W = parenleftbigg 2 3 - 3 2 parenrightbigg U out 8. W = parenleftbigg 2 3- 3 2 parenrightbigg U out 9. W = ( - 1) U out correct 10. W = (1- ) U out Explanation: The capacitance of a capacitor with a di- electric slab is C in = C out , where > 1 , and U out = 1 2 Q out V . NOTE When the battery remains connected, the electric potential difference across the plates of the capacitor will remain constant; i.e. , V in = V out = V = constant. The difference in the potential energy stored in the capacitor is U cap = 1 2 Q in V- 1 2 Q out V = 1 2 Q out V- 1 2 Q out V = ( - 1) parenleftbigg 1 2 Q out V parenrightbigg , and the energy drained from the battery is U bat = ( Q out- Q in ) V = (1- ) Q out V , so the total energy difference U total is U in- U out = ( - 1) parenleftbigg 1 2 Q out V parenrightbigg + (1- ) Q out V = parenleftbigg 1 2 Q out- 1 2 Q out parenrightbigg V = (1- ) 1 2 Q out V = (1- ) U out , since W =- U total W = ( - 1) U out , where U out is with an air-filled gap and U in is with a dielectric-filled gap. It takes work to move the dielectric out of the capacitor and U out > U in . A system will move to a position of lower potential energy. If the dielectric is moved half way out of the capacitor, the potential energy stored in the soto (rrs946) Homework 6 Hoffmann (56605) 2 capacitor will be smaller than it would have been with the dielectric left in place; however, the energy increase in charging the battery will be twice as large as the energy loss in the...
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Homework 6-solutions - soto (rrs946) Homework 6 Hoffmann...

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