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Unformatted text preview: homework 04 – HOFFMA, DAVID – Due: Feb 11 2008, 4:00 am 1 Question 1, chap 26, sect 1. part 1 of 1 10 points Two large parallel conducting plates P and Q are connected to a battery of emf E , as shown. A test charge is placed successively at points I , II , and III . E + I II III If edge effects are negligible, the force on the charge when it is at point III is 1. much greater in magnitude than the force on the charge when it is at point II , but in the same direction. 2. of equal magnitude and in the same di rection as the force on the charge when it is at point I . 3. of equal magnitude and in the same di rection as the force on the charge when it is at point II . correct 4. much less in magnitude than the force on the charge when it is at point II , but in the same direction. 5. equal in magnitude to the force on the charge when it is at point I , but in the oppo site direction. Explanation: Neglecting edge effects, the electric field strength between the two plates is uniform. Therefore the force on the test charge is the same at points II and III , both in magnitude and in direction. When edge effects are negligible, the elec tric field strength at point I is zero, so the force on the test charge is zero when it is at point I . Question 2, chap 26, sect 1. part 1 of 1 10 points A spherical capacitor consists of a conduct ing ball of radius 13 cm that is centered inside a grounded conducting spherical shell of inner radius 15 cm. The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . What charge is required to achieve a poten tial of 1574 V on the ball of radius 13 cm? Correct answer: 1 . 70753 × 10 − 7 C (tolerance ± 1 %). Explanation: Let : k e = 8 . 98755 × 10 9 N m 2 / C 2 , a = 13 cm = 0 . 13 m , b = 15 cm = 0 . 15 m , and V = 1574 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 . 13 m) (0 . 15 m) (1574 V) (0 . 15 m) (0 . 13 m) = 1 . 70753 × 10 − 7 C . Question 3, chap 26, sect 1. part 1 of 1 10 points A small electrically charged object is sus pended by a thread between the vertical plates of a parallelplate capacitor. The acceleration of gravity is 9 . 8 m / s 2 . homework 04 – HOFFMA, DAVID – Due: Feb 11 2008, 4:00 am 2 A B vector E + E 215 mg 36 nC θ θ =20 ◦ 4 . 6 cm What is the potential difference between the plates? Correct answer: 0 . 979909 kV (tolerance ± 1 %). Explanation: Let : m = 215 mg = 0 . 000215 kg , g = 9 . 8 m / s 2 , d = 4 . 6 cm = 0 . 046 m , θ = 20 ◦ , and q = 36 nC = 3 . 6 × 10 − 8 C ....
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This note was uploaded on 03/19/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Charge, Work

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