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PHYpractice1

# PHYpractice1 - Howard(clh2528 – Practice problems on...

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Unformatted text preview: Howard (clh2528) – Practice problems on chapters 23 and 24 – shubeita – (57885) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 6 . 34 cm 2 , sepa- rated by a distance of 2 . 77 mm. A(n) 29 . 9 V potential difference is applied to these plates. Find the magnitude of the electric field be- tween the plates. Correct answer: 10 . 7942 kV / m. Explanation: Let : A = 6 . 34 cm 2 = 0 . 000634 m 2 , d = 2 . 77 mm = 0 . 00277 m , V = 29 . 9 V , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . The electric field is E = Δ V d = 29 . 9 V . 00277 m × 1 kV 1000 V = 10 . 7942 kV / m directed toward the negative plate. 002 (part 2 of 3) 10.0 points Find the capacitance. Correct answer: 2 . 02655 pF. Explanation: C = ǫ A d = (8 . 85419 × 10 − 12 C 2 / N · m 2 ) × parenleftbigg . 000634 m 2 . 00277 m parenrightbigg × parenleftbigg 10 12 pF 1 F parenrightbigg = 2 . 02655 pF . 003 (part 3 of 3) 10.0 points Find the magnitude of the charge on each plate. Correct answer: 60 . 594 pC. Explanation: The charge is Q = C V = (2 . 02655 pF) (29 . 9 V) = 60 . 594 pC . on one plate and- 60 . 594 pC on the other. 004 10.0 points A parallel-plate capacitor is charged by con- necting it to a battery. If 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 decreases and the electric po- tential increases. 3. The charge remains fixed and the electric potential increases. correct 4. The charge increases and the electric po- tential remains fixed. 5. The charge decreases and the electric po- tential remains fixed. 6. The charge and the electric potential de- crease. 7. The charge and the electric potential re- main fixed. 8. The charge remains fixed and the electric potential decreases. 9. The charge increases and the electric po- tential decreases. Explanation: Howard (clh2528) – Practice problems on chapters 23 and 24 – shubeita – (57885) 2 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 A d A larger d makes the fraction smaller, so C is smaller. Thus the new potential V ′ = Q C ′ is larger. 005 (part 1 of 2) 10.0 points Consider the system of capacitors connected to a battery as shown below. E B 2 . 3 μ F 2 . 3 μ F 2 . 3 μ F 2 . 3 μ F 2 . 3 μ F 2 . 3 μ F Find the equivalent capacitance of the cir- cuit....
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PHYpractice1 - Howard(clh2528 – Practice problems on...

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