1178864680Xuj

1178864680Xuj - Alvarado, Patrick Homework 5 Due: Feb 23...

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Alvarado, Patrick – Homework 5 – Due: Feb 23 2007, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points A parallel-plate capacitor is charged by con- necting it to a battery. I± the battery is disconnected and the sep- aration between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it? 1. The charge and the electric potential in- crease. 2. The charge and the electric potential re- main fxed. 3. The charge increases and the electric po- tential decreases. 4. The charge remains fxed and the electric potential increases. correct 5. The charge remains fxed and the electric potential decreases. 6. The charge decreases and the electric po- tential increases. 7. The charge decreases and the electric po- tential remains fxed. 8. The charge and the electric potential de- crease. 9. The charge increases and the electric po- tential remains fxed. Explanation: Charge is conserved, so it must remain con- stant since it is stuck on the plates. With the battery disconnected, Q is fxed. C = ² A d A larger d makes the ±raction smaller, so C is smaller. Thus the new potential V 0 = Q C 0 is larger. keywords: 002 (part 1 o± 1) 10 points When a potential di²erence o± 230 V is ap- plied to the plates o± a parallel-plate capaci- tor, the plates carry a sur±ace charge density o± 18 nC / cm 2 . The permittivity o± a vacuum is 8 . 85419 × 10 - 12 C 2 / N · m 2 . What is the spacing between the plates? Correct answer: 11 . 3137 μ m. Explanation: Let : σ = 18 nC / cm 2 = 0 . 00018 C / m 2 , V = 230 V , and ² 0 = 8 . 85419 × 10 - 12 C 2 / N · m 2 . s = ² 0 V σ = (8 . 85419 × 10 - 12 C 2 / N · m 2 ) (230 V) (0 . 00018 C / m 2 ) = 1 . 13137 × 10 - 5 m = 11 . 3137 μ m . keywords: 003 (part 1 o± 2) 10 points A parallel-plate capacitor has a capacitance o± 1 μ F and a plate separation o± 4 mm. The dielectric strength the air is 3 × 10 6 V / m. What is the maximum potential di²erence between the plates, so that dielectric break- down o± the air between the plates does not occur? Correct answer: 12 kV. Explanation: Let : d = 4 mm = 0 . 004 m , and E m = 3 × 10 6 V / m .
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Alvarado, Patrick – Homework 5 – Due: Feb 23 2007, 11:00 pm – Inst: Opyrchal, H 2 The maximum potential diference is V m = E m d = (3 × 10 6 V / m) (0 . 004 m) · kV 10 3 V = 12 kV .
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1178864680Xuj - Alvarado, Patrick Homework 5 Due: Feb 23...

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