18_InstSolManual_PDF_Part1

18_InstSolManual_PDF_Part1 - ELECTRIC POTENTIAL AND...

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Unformatted text preview: ELECTRIC POTENTIAL AND CAPACITANCE 18 9. B 10. C 11. D 12. A 13. A 14. A, C 15. C Answers to Multiple-Choice Problems 1. A, C 2. C 3. A 4. A, D 5. C 6. B, C 7. B, D 8. C Solutions to Problems 18.1. Set Up: Since the charge is positive the force on it is in the same direction as the electric field. Since the field is uniform the force is constant and W 5 Fs cos f. S S Solve: (a) F is upward and s is to the right, so f 5 90°and W 5 0. S S (b) F is upward and s is upward, so f 5 0°. S W 5 Fs 5 qEs 5 1 28.0 3 1029 C 2 1 4.00 3 104 N / C 2 1 0.670 m 2 5 7.50 3 1024 J. S (c) F is upward and s is at 45.0° below the horizontal, so f 5 135.0°. W 5 Fs cos f 5 qEs cos f 5 1 28.0 3 1029 C 2 1 4.00 3 104 N / C 2 1 2.60 m 2 cos 135.0° 5 22.06 3 1023 J. Reflect: The work is positive when the displacement has a component in the direction of the force and it is negative when the displacement has a component opposite to the direction of the force. When the displacement is perpendicular to the force the work done is zero. 18.2. Set Up: (a) The proton has charge 1e and mass 1.67 3 10227 kg. Let point a be at the negative plate and point b be at the positive plate. The electric field is directed from the positive plate toward the negative plate. The force on the positively charged proton is in this direction. The proton moves in a direction opposite to the electric force, so the work done by the electric force is negative. Solve: Wa S b 5 2qEd 5 2eEd 5 2 1 1.60 3 10219 C 2 1 2.80 3 106 N / C 2 1 0.100 m 2 5 24.48 3 10214 J q1q2 r 1 8.99 3 109 N # m2 / C2 2 1 27.20 3 1026 C 2 112.30 3 1026 C 2 kq1q2 Solve: r 5 5 5 0.372 m U 20.400 J 18.3. Set Up: U 5 k q1q2 . ra 5 0.150 m, rb 5 " 1 0.250 m 2 2 1 1 0.250 m 2 2 5 0.354 m. r 1 8.99 3 109 N # m2 / C2 2 1 2.40 3 1026 C 2 1 24.30 3 1026 C 2 q1q2 Solve: Ua 5 k 5 5 20.619 J. ra 0.150 m 18.4. Set Up: Wa S b 5 Ua 2 Ub . U 5 k Ub 5 k 1 8.99 3 109 N # m2 / C2 2 1 2.40 3 1026 C 2 1 24.30 3 1026 C 2 q1q2 5 5 20.262 J. rb 0.354 m Wa S b 5 Ua 2 Ub 5 20.619 J 2 1 20.262 J 2 5 20.357 J. 18-1 ...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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