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Lecture 9 jan31 - Lecture#9 Energy storage in Capacitors...

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Exam Locations: Vallery (7, 8, 9): AUD B Angell Kurdak (10,11,12): 182 Dennison Krisch (13,14,15): 170 Dennison Lecture #9 Energy storage in Capacitors Dielectrics Good luck on the exam tonight! Bring pencils, note card, calculator, student ID “Is that Your Final Answer?”-----the Scantron always is. Results should be on SAMS this weekend
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Energy “stored” in a capacitor (Demo: The “Killer Capacitor”) Separating + from - charge requires work by an external force: Moving charge Q through ΔV =V requires ΔU = W ext = QV (NOT HERE!) (This is indeed the work done by the battery!) But V of capacitor builds up as q builds up: V = q/C Thus, dU = Vdq = dW ext = (q/C)dq 2 2 0 2 0 2 1 2 1 2 1 2 1 CV QV C Q C q C qdq U Q Q q = = = = = = So why the factor of ½ here??? Question #1>>>>>
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Question #1 The work done by the battery to transfer Q from one conductor of the capacitor to the other is QV. The energy stored in the capacitor is only (½)QV. So what happened to the other (1/2)QV of energy?
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