homework 11-solutions

homework 11-solutions - howard(cah3459 homework 11...

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howard (cah3459) – homework 11 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A parallel-plate capacitor oF dimensions 2 . 39 cm × 5 . 77 cm is separated by a 1 . 18 mm thickness oF paper. ±ind the capacitance oF this device. The dielectric constant κ For paper is 3.7. Correct answer: 38 . 2862 p±. Explanation: Let : κ =3 . 7 , d =1 . 18 mm = 0 . 00118 m , and A =2 . 39 cm × 5 . 77 cm = 0 . 00137903 m 2 . We apply the equation For the capacitance oF a parallel-plate capacitor and fnd C = κ! 0 A d = (3 . 7) (8 . 85419 × 10 - 12 C 2 / N · m 2 ) × ± 0 . 00137903 m 2 0 . 00118 m ² 1 × 10 12 = 38 . 2862 p± . 002 (part 2 oF 2) 10.0 points What is the maximum charge that can be placed on the capacitor? The electric strength oF paper is 1 . 6 × 10 7 V / m. Correct answer: 0 . 722843 μ c. Explanation: Let : E max . 6 × 10 7 V / m . Since the thickness oF the paper is 0 . 00118 m, the maximum voltage that can be applied beFore breakdown is V max = E max d. Hence, the maximum charge is Q max = CV max = CE max d = (38 . 2862 p±)(18880 V) · 1 × 10 - 12 ± 1 p± · 1 × 10 6 μ C 1C = 0 . 722843 μ c . 003 (part 1 oF 4) 10.0 points Determine the total energy stored in a con- ducting sphere with charge Q . Hint: Use the capacitance Formula For a spherical capacitor which consists oF two spherical shells. Take the inner sphere to have a radius a and the outer shell to have an infnite radius. 1. U = Q 2 a 4 π! 0 2. U = Q 2 π 8 ! 0 a 3. U = Q 8 0 a 4. U = Q 2 16 0 a 5. U = Q 2 a 6. U = Q 2 8 0 a 2 7. U = Q 2 8 0 a correct 8. U = Q 2 4 0 a Explanation: The capacitance Formula For a spherical ca- pacitor oF inner radius a and outer radius b is C = ab k e ( b - a ) . IF we let b →∞ , we fnd we can neglect a in the denominator compared to b , so C a k e =4 0 a. The total energy stored is U = Q 2 2 C = Q 2 8 0 a .
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howard (cah3459) – homework 11 – turner – (56705) 2 004 (part 2 of 4) 10.0 points Find the energy stored in a capacitor of charge Q ±lled with dielectric, use C κ = κ C . 1. U = Q 2 3( κ - 1) C 2. U = Q 4 3. U = Q 2 3 C 4. U = Q 2 5. U = Q 2 2 correct 6. U = Q 2 7. U = Q 2 3 8. U = Q 2 2( κ - 1) C 9. U = Q 2 2 C Explanation: The energy stored in a capacitor of capaci- tance C κ is U = Q 2 2 C κ = Q 2 2 . 005 (part 3 of 4) 10.0 points Work = U f - U i , where “i” is the initial state where there is a slab in the gap and “f” is the ±nal state where there is no slab in the gap. Find the work done in pulling a dielectric slab of dielectric constant κ from the gap of a parallel plate capacitor of plate charge Q and capacitance C . 1. W if = Q 2 2 C 2. W if = Q 2 2 C ± κ + 1 κ ² 3. W if = Q 2 2 4. W if = Q 2 2 C ± 1 - 1 κ ² correct 5. W if = Q 2 2 C ± 1 κ - 1 ² 6. W if = Q 2 2 C ± κ - 1 κ ² 7. W if = Q 2 5 8. W if = Q 2 2 (1 - κ ) 9. W if = Q 2 2 C (1 - κ ) Explanation: The energy stored in the capacitor with the dielectric is U i = Q 2 2 . After pulling the slab out, the charge on the capacitor remains the same (charge conserva- tion). The energy stored is U f = Q 2 2 C , so the work that must be done to pull the slab out is W = U f - U i = Q 2 2 C - Q 2 2 , or W = Q 2 2 C ± 1 - 1 κ ² .
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homework 11-solutions - howard(cah3459 homework 11...

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