# Ch21Word - Chapter 21 p. 1 CHAPTER 21 – Electric Charge...

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Unformatted text preview: Chapter 21 p. 1 CHAPTER 21 – Electric Charge and Electric Field 1. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 = ( 9 . × 10 9 N · m 2 /C 2 )(2.50 C)(2.50 C)/(3.0 m) 2 = 6.3 × 10 9 N . 2. The number of electrons is N = Q /(– e ) = (– 30.0 × 10 –6 C)/(– 1.60 × 10 –19 C/electrons) = 1.88 × 10 14 electrons . 3. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 = ( 9 . × 10 9 N · m 2 /C 2 )(26)(1.60 × 10 –19 C)(1.60 × 10 –19 C)/(1.5 × 10 –12 m) 2 = 2.7 × 10 –3 N . 4. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 = ( 9 . × 10 9 N · m 2 /C 2 )(1.60 × 10 –19 C)(1.60 × 10 –19 C)/(5.0 × 10 –15 m) 2 = 9.2 N . 5. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 = ( 9 . × 10 9 N · m 2 /C 2 )(25 × 10 –6 C)(3.0 × 10 –3 C)/(0.35 m) 2 = 5.5 × 10 3 N . 6. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 . If we divide the expressions for the two forces, we have F 2 / F 1 = ( r 1 / r 2 ) 2 ; F 2 /(4.2 × 10 –2 N) = (8) 2 , which gives F 2 = 2.7 N . 7. The magnitude of the Coulomb force is F = kQ 1 Q 2 / r 2 . If we divide the expressions for the two forces, we have F 2 / F 1 = ( r 1 / r 2 ) 2 ; 3 = [(15.0 cm)/ r 2 ] 2 , which gives r 2 = 8.66 cm . 8. The number of excess electrons is N = Q /(– e ) = (– 40 × 10 –6 C)/(– 1.60 × 10 –19 C/electrons) = 2.5 × 10 14 electrons . The mass increase is ? m = Nm e = (2.5 × 10 14 electrons)(9.11 × 10 –31 kg/electron) = 2.3 × 10 –16 kg . 9. The number of molecules in 1.0 kg H 2 O is N = [(1.0 kg)(10 3 g/kg)/(18 g/mol)](6.02 × 10 23 molecules/mol) = 3.34 × 10 25 molecules. Each molecule of H 2 O contains 2(1) + 8 = 10 electrons. The charge of the electrons in 1.0 kg is q = (3.34 × 10 25 molecules)(10 electrons/molecule)(– 1.60 × 10 –19 C/electron) = – 5.4 × 10 7 C . Chapter 21 p. 2 10. Using the symbols in the figure, we find the magnitudes of the three individual forces: F 12 = F 21 = kQ 1 Q 2 / r 12 2 = kQ 1 Q 2 / L 2 = ( 9 . × 10 9 N · m 2 /C 2 )(70 × 10 –6 C)(48 × 10 –6 C)/(0.35 m) 2 = 2.47 × 10 2 N. F 13 = F 31 = kQ 1 Q 3 / r 13 2 = kQ 1 Q 3 /(2 L ) 2 = ( 9 . × 10 9 N · m 2 /C 2 )(70 × 10 –6 C)(80 × 10 –6 C)/[2(0.35 m)] 2 = 1.03 × 10 2 N. F 23 = F 32 = kQ 2 Q 3 / r 23 2 = kQ 2 Q 3 / L 2 = ( 9 . × 10 9 N · m 2 /C 2 )(48 × 10 –6 C)(80 × 10 –6 C)/(0.35 m) 2 = 2.82 × 10 2 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the net forces, we get F 1 = F 13 – F 12 = 1.03 × 10 2 N – 2.47 × 10 2 N = – 1.4 × 10 2 N (left). F 2 = F 21 + F 23 = 2.47 × 10 2 N + 2.82 × 10 2 N = + 5.3 × 10 2 N (right). F 3 = – F 31 – F 32 = – 1.03 × 10 2 N – 2.82 × 10 2 N = – 3.9 × 10 2 N (left)....
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch21Word - Chapter 21 p. 1 CHAPTER 21 – Electric Charge...

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