Homework 5-solutions - kumar(vk4433 Homework 5...

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kumar (vk4433) – Homework 5 – markert – (55325) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A 14 pF capacitor is connected across a 56 V source. What charge is stored on it? Correct answer: 7 . 84 × 10 10 C. Explanation: Let : C = 14 pF = 1 . 4 × 10 11 F and V = 56 V . The capacitance is C = q V q = C V = (1 . 4 × 10 11 F) (56 V) = 7 . 84 × 10 10 C . 002 10.0points Two conductors not touching each other are charged by transferring electrons from one conductor to the other. After 1 . 7022 × 10 14 have been transferred, the potential difference between the conductors is 7 . 5 V. What is the capacitance of the system? The fundamental charge is 1 . 602 × 10 19 C . Correct answer: 3 . 6359 × 10 6 F. Explanation: Let : e = 1 . 602 × 10 19 C , N = 1 . 7022 × 10 14 electrons , and V = 7 . 5 V . One conductor gains a charge of Q = N e while the other gains a charge of Q , so the capacitance of the system is C = q V = N e V = (1 . 7022 × 10 14 ) (1 . 602 × 10 19 C) 7 . 5 V = 3 . 6359 × 10 6 F . 003 10.0points A parallel-plate capacitor is charged by con- necting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capac- itor and the electric potential across it? 1. The charge increases and the electric po- tential remains fixed. 2. The charge decreases and the electric po- tential remains fixed. 3. The charge and the electric potential re- main fixed. 4. The charge remains fixed and the electric potential increases. correct 5. The charge remains fixed and the electric potential decreases. 6. The charge decreases and the electric po- tential increases. 7. The charge and the electric potential de- crease. 8. The charge and the electric potential in- crease. 9. The charge increases and the electric po- tential decreases. Explanation: Charge is conserved, so it must remain con- stant since it is stuck on the plates. With the battery disconnected, Q is fixed: C = ǫ 0 A d . A larger d makes the fraction smaller, so C is smaller, making the new potential V = Q C larger.
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kumar (vk4433) – Homework 5 – markert – (55325) 2 004 10.0points A circular parallel-plate capacitor with a spacing of 3 . 5 mm is charged to produce a uniform electric field with a strength of 2 . 7 × 10 6 N / C. What plate radius is required if the stored charge is 1 . 06 μ C? The permittivity of a vacuum is 8 . 8542 × 10 12 C 2 / N · m 2 . Correct answer: 0 . 118801 m. Explanation: Let : Δ d = d = 3 . 5 mm , Q = 1 . 06 μ C = 1 . 06 × 10 6 C , E = 2 . 7 × 10 6 N / C , and ǫ 0 = 8 . 8542 × 10 12 C 2 / N · m 2 . The capacitance is C = ǫ 0 A d = Q Δ V ǫ 0 π r 2 d = Q E Δ d r = radicalbigg Q π ǫ 0 E = radicalBigg 1 . 06 × 10 6 C π (8 . 8542 × 10 12 C 2 / N · m 2 ) × radicalBigg 1 2 . 7 × 10 6 N / C = 0 . 118801 m . 005 10.0points A 1-megabit computer memory chip contains many 7 . 3 × 10 14 F capacitors. Each capaci- tor has a plate area of 3 × 10 11 m 2 .
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