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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

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.

Page1 / 26

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online