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Unformatted text preview: homework 04 – BOBBITT, JAMES – 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 II . correct 3. of equal magnitude and in the same di rection as the force on the charge when it is at point I . 4. equal in magnitude to the force on the charge when it is at point I , but in the oppo site direction. 5. much less in magnitude than the force on the charge when it is at point II , but in the same 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 4 cm that is centered inside a grounded conducting spherical shell of inner radius 6 cm. The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . What charge is required to achieve a poten tial of 1088 V on the ball of radius 4 cm? Correct answer: 1 . 45268 × 10 − 8 C (tolerance ± 1 %). Explanation: Let : k e = 8 . 98755 × 10 9 N m 2 / C 2 , a = 4 cm = 0 . 04 m , b = 6 cm = 0 . 06 m , and V = 1088 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 . 04 m) (0 . 06 m) (1088 V) (0 . 06 m) (0 . 04 m) = 1 . 45268 × 10 − 8 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 – BOBBITT, JAMES – Due: Feb 11 2008, 4:00 am 2 A B vector E + E 372 mg 37 nC θ θ =20 ◦ 5 . 5 cm What is the potential difference between the plates? Correct answer: 1 . 9724 kV (tolerance ± 1 %). Explanation: Let : m = 372 mg = 0 . 000372 kg , g = 9 . 8 m / s 2 , d = 5 . 5 cm = 0 . 055 m , θ = 20 ◦ , and q = 37 nC = 3 . 7 × 10 − 8 C ....
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This note was uploaded on 09/18/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Charge, Work

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