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Homework 6-solutions - gonzales(pag757 Homework 6...

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gonzales (pag757) – Homework 6 – Hoffmann – (56745) 1 This print-out should have 12 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 Q V = 1 2 C V 2 . a) An isolated capacitor has a dielectric slab κ between its plates. b) The capacitor is charged by a battery. c) After the capacitor is fully charged, the battery is removed. d) Finally, the dielectric slab is moved out of the capacitor. κ κ 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 = parenleftbigg 2 3 κ - 3 2 parenrightbigg U out 2. W = ( κ - 1) U out 3. W = 3 2 ( κ - 1) U out 4. W = parenleftbigg 1 κ - 1 parenrightbigg U out 5. W = parenleftbigg 3 2 κ - 2 3 parenrightbigg U out 6. W = (1 - κ ) U out 7. W = parenleftbigg 2 3 - 3 2 κ parenrightbigg U out 8. W = parenleftbigg 1 - 1 κ parenrightbigg U out correct 9. W = 3 2 parenleftbigg 1 - 1 κ parenrightbigg U out 10. W = parenleftbigg 3 2 - 2 3 κ parenrightbigg U out Explanation: The capacitance of a capacitor with a di- electric slab is C in = κ C out , where κ > 1 , and U out = 1 2 V 2 C out . NOTE When the battery is removed, the charge on the plates of the capacitor will remain constant. Charge is neither created nor de- stroyed; i.e. , Q in = Q out = Q = constant. The difference in the potential energy stored in the capacitor is Δ U cap = 1 2 Q 2 C in - 1 2 Q 2 C out = 1 2 Q 2 κ C in - 1 2 Q 2 C out = parenleftbigg 1 κ - 1 parenrightbigg 1 2 Q 2 C out , and the energy drained from the battery is Δ U bat = ( Q out - Q in ) V = 0 , since Δ Q = 0 , so the total energy difference Δ U total is U in - U out = parenleftbigg 1 κ - 1 parenrightbigg 1 2 Q 2 C out = parenleftbigg 1 κ - 1 parenrightbigg U out , since W = - Δ U total W = parenleftbigg 1 - 1 κ parenrightbigg 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 capacitor will be larger than it would have been with the dielectric left in place.
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gonzales (pag757) – Homework 6 – Hoffmann – (56745) 2 Therefore, the dielectric will be pulled back into the capacitor. 002 10.0 points A capacitor is constructed from two metal plates. The bottom portion is filled with air and the top portion is filled with material of dielectric constant κ . The plate area in the top region is the same as that in the bottom region. Neglect edge effects. κ E Determine the ratio U t U b of energy stored in the top portion to the bottom portion.
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