{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

681_PartUniversity Physics Solution

681_PartUniversity Physics Solution - 23-4 Chapter 23(iii...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
23-4 Chapter 23 (iii) 50.0 m b r = b r is now ten times larger than in (ii) so U b is ten times smaller: 0.00992 J/10 0.000992 J. b U = + = + 4 2( ) 2( 0.198 J 0.000992 J) 37.5 m/s. 2.80 10 kg a b b U U v m + = = = × E VALUATE : The force between the two charges is repulsive and provides an acceleration to q. This causes the speed of q to increase as it moves away from Q. 23.8. I DENTIFY : Call the three charges 1, 2 and 3. 12 13 23 U U U U = + + S ET U P : 12 23 13 U U U = = because the charges are equal and each pair of charges has the same separation, 0.500 m. E XECUTE : 2 6 2 3 3 (1.2 10 C) 0.078 J. 0.500 m 0.500 m kq k U × = = = E VALUATE : When the three charges are brought in from infinity to the corners of the triangle, the repulsive electrical forces between each pair of charges do negative work and electrical potential energy is stored. 23.9. I DENTIFY : 1 2 1 3 2 3 12 13 23 q q q q q q U k r r r = + + S ET U P : In part (a), 12 0.200 m r = , 23 0.100 m r = and 13 0.100 m. r = In part (b) let particle 3 have coordinate x , so 12 0.200 m r = , 13 r x = and 23 0.200 . r x = E XECUTE : (a) 7 (4.00 nC)( 3.00 nC) (4.00 nC)(2.00 nC) ( 3.00 nC)(2.00 nC) 3.60 10 J (0.200 m) (0.100 m) (0.100 m) U k = + + = × (b) If 0 U = , then 1 2 1 3 2 3 12 12 0 . q q q q q q k r x r x = + + Solving for x we find: 2 8 6 0 60 60 26 1.6 0 0.074 m, 0.360 m. 0.2 x x x x x = − + + = = Therefore, 0.074 m x = since it is the only value between the two charges. E VALUATE : 13 U is positive and both 23 U and 12 U are negative. If 0 U = , then 13 23 12 . U U U = + For 0.074 m x = , 7 13 9.7 10 J U = + × , 7 23 4.3 10 J U = − × and 7 12 5.4 10 J. U = − × It is true that 0 U = at this x . 23.10. I DENTIFY : The work done on the alpha particle is equal to the difference in its potential energy when it is moved from the midpoint of the square to the midpoint of one of the sides. S ET U P : We apply the formula . a b a b W U U = In this case, a is the center of the square and b is the midpoint of one of the sides. Therefore center side center side . W U U = There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single alpha-electron pair. At the center of the square, the alpha particle is a distance r 1 = 50 nm from each electron. At the midpoint of the side, the alpha is a distance r 2 = 5.00 nm from the two nearest electrons and a distance r 2 = 125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of two point charges, 0 0 (1/4 )( / ), U qq r π = P the total work is center side center side W U U = = 0 1 0 2 0 3 1 1 1 4 2 2 4 4 4 e e e q q q q q q r r r α α α π π π + P P P Substituting q e = e and q α = 2 e and simplifying gives 2 center side 0 1 2 3 1 2 1 1 4 4 W e r r r π = − + P E XECUTE : Substituting the numerical values into the equation for the work gives ( ) 2 19 21 2 1 1 4 1.60 10 C 6.08 10 J 5.00 nm 50 m 125 nm W = − × + = × ???
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern