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# HW4 - oliver(jmo673 HW04 Tsoi(59090 This print-out should...

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oliver (jmo673) – HW04 – Tsoi – (59090) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 9 . 6 cm 2 , sepa- rated by a distance 3 . 5 mm . A 21 V potential difference is applied to these plates. The permittivity of a vacuum is 8 . 85419 × 10 12 C 2 / N · m 2 . 1 pF is equal to 10 12 F . The magnitude of the electric field between the plates is 1. E = ( V d ) 2 . 2. E = parenleftbigg V d parenrightbigg 2 . 3. None of these 4. E = V d . correct 5. E = V d . 6. E = parenleftbigg d V parenrightbigg 2 . 7. E = 1 ( V d ) 2 . 8. E = 1 V d . 9. E = d V . Explanation: Since E is constant between the plates, V = integraldisplay vector E · d vector l = E d E = V d . 002 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. None of these 2. σ = ǫ 0 parenleftbigg V d parenrightbigg 2 . 3. σ = ǫ 0 d V . 4. σ = ǫ 0 V d . 5. σ = ǫ 0 ( V d ) 2 . 6. σ = ǫ 0 ( V d ) 2 . 7. σ = ǫ 0 V d 8. σ = ǫ 0 V d . correct 9. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 . Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ǫ 0 E = ǫ 0 V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it. 003 (part 3 of 4) 10.0 points

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oliver (jmo673) – HW04 – Tsoi – (59090) 2 Calculate the capacitance. Correct answer: 2 . 42858 pF. Explanation: Let : A = 0 . 00096 m 2 , d = 0 . 0035 m , V = 21 V , and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . The capacitance is given by C = ǫ 0 A d = 8 . 85419 × 10 12 C 2 / N · m 2 × 0 . 00096 m 2 0 . 0035 m = 2 . 42858 × 10 12 F = 2 . 42858 pF . 004 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 51 . 0001 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (2 . 42858 × 10 12 F) (21 V) = 5 . 10001 × 10 11 C = 51 . 0001 pC . 005 10.0 points A spherical capacitor consists of a conducting ball of radius 16 cm that is centered inside a grounded conducting spherical shell of inner radius 19 cm. What charge is required to achieve a potential of 1224 V on the ball of ra- dius 16 cm? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 38004 × 10 7 C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , a = 16 cm = 0 . 16 m , b = 19 cm = 0 . 19 m and V = 1224 V . The capacitance of a spherical capacitor is given by C = a b k e ( b - a ) . The charge required to achieve a potential difference V between the shell and the ball is given by Q = C V = 1 k e a b V ( b - a ) = 1 (8 . 98755 × 10 9 N · m 2 / C 2 ) × (0 . 16 m) (0 . 19 m) (1224 V) (0 . 19 m) - (0 . 16 m) = 1 . 38004 × 10 7 C . keywords: 006 (part 1 of 2) 10.0 points A capacitor network is shown below. 112 V 16 μ F 16 μ F 16 μ F 16 μ F 14 μ F 16 μ F 16 μ F y z What is the equivalent capacitance between points y and z of the entire capacitor net- work?
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HW4 - oliver(jmo673 HW04 Tsoi(59090 This print-out should...

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