HW11Sol - Platt David Homework 11 Due Oct 3 2005 4:00 am...

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Platt, David – Homework 11 – Due: Oct 3 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 2) 10 points Consider an air-flled parallel plate capaci- tor with plate area A and gap width d . The plate charge is Q . E d + Q - Q A A The total energy stored in the capacitor is given by 1. U = Q 2 ² 0 Ad . 2. U = Q 2 d 2 ² 0 A . correct 3. U = Qd ² 0 A . 4. U = Q 2 A ² 0 d . 5. U = QA ² 0 d . 6. U = Q 2 ² 0 . 7. U = Q ² 0 . 8. U = Q 2 2 ² 0 . Explanation: U = F d = Q 2 2 ² 0 A d. Alternate Solution: U = Q 2 2 C = Q 2 2 µ ² 0 A d = Q 2 2 ² 0 A 002 (part 2 oF 2) 10 points With the battery connected, fll the gap by a slab with the dielectric constant κ . Compare the new plate charge Q 0 with Q , the plate charge in part 1. Choose one: 1. Q 0 = 2 Q κ - 1 2. Q 0 = κQ correct 3. Q 0 = ( κ + 1) Q 4. Q 0 = κ - 1 2 Q 5. Q 0 = Q κ + 1 6. Q 0 = ( κ - 1) Q 7. Q 0 = Q κ - 1 8. Q 0 = 2 Q κ + 1 9. Q 0 = κ + 1 2 Q 10. Q 0 = Q κ Explanation: Q 0 = E C 0 = E κC = κQ. 003 (part 1 oF 2) 10 points Consider the setup shown, where a capacitor with a capacitance C is connected to a battery with emF V and negligible internal resistance. BeFore the insertion oF the dielectric slab with dielectric constant κ , the charge on the capacitor is Q = C V and the energy density is u = 1 2 ² 0 E 2 . Now, keeping the battery connected, insert the dielectric, which flls the gap completely.
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Platt, David – Homework 11 – Due: Oct 3 2005, 4:00 am – Inst: Ken Shih 2 d V C Find the plate charge Q 0 plate .
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HW11Sol - Platt David Homework 11 Due Oct 3 2005 4:00 am...

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