chap20 - ELECTRIC CHARGE, FORCE, AND FIELD 20 EXERCISES...

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20.1 ELECTRIC CHARGE, FORCE, AND FIELD 20 EXERCISES Section 20.1 Electric Charge 14. Nearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in living matter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg average-sized person is approximately 27 28 1 2 (65 kg)/(1.67 10 kg) 2 10 , × < × 0 , e which is also the number of electrons, since an average person is electrically neutral. If there were a charge imbalance of a person’s net charge would be about or several coulombs (huge by ordinary standards). 9 proton electron || 1 qq −= 28 9 19 2 10 10 1.6 10 C 3.2 C, −− ±× × × × 15. INTERPRET This problem deals with quantity of charge in a typical lightning flash. We want to express the quantity in terms of elementary charge e . DEVELOP Since the magnitude of elementary charge e is the number of electrons involved is given by 19 1.6 10 C, e /. Qe EVALUATE Substituting the values given in the problem statement, we find the number to be 20 19 25 C /1 . 1.6 10 C NQ e == × 5 6 1 0 ASSESS Since 1 coulomb is about elementary charges, our result has the right order of magnitude. 18 6.25 10 × 16. (a) The proton’s charge is 221 333 1 corresponding to a combination of uud quarks; (b) for neutrons, , eeee =+− 0 = 211 corresponds to udd . (See Chapter 39, or Chapter 45 in the extended version of the text.) Section 20.2 Coulomb’s Law 17. INTERPRET In this problem we are asked to compare the gravitational and electrical forces between a proton and an electron. DEVELOP The gravitational force and the electrostatic force between a proton and an electron separated by a distance r are, respectively, grav 2 pe Gm m F r = and 2 elec 2 ke F r = EVALUATE The ratio of the two forces is 22 9 2 2 1 9 2 elec 21 1 2 2 2 7 grav 39 (9 10 N m /C )(1.6 10 C) (6.67 10 N m /kg )(1.67 10 kg)(9.11 10 kg) 2.3 10 F ke r FG m m r ⎛⎞ ×⋅ × ⎜⎟ × × ⎝⎠ ≈× 3 1 Note that the spatial dependence of both forces is the same, and cancels out. ASSESS At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is about10 times stronger than the gravitational force. 40
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20.2 Chapter 20 18. pm is called the Bohr radius. For a proton and electron separated by a Bohr radius, 0 52.9 a = 22 Coulomb 0 / Fk e a = . 92 2 1 9 1 1 2 8 (9 10 N m /C )(1.6 10 C/5.29 10 m) 8.23 10 N. −− ×⋅ × × = × 19. INTERPRET This problem is about comparing the gravitational and electrical forces. DEVELOP The electric force between a proton and an electron has magnitude elec /, F ke r = while the weight of an electron is . ge F mg = EVALUATE When the two forces are equal, elec , g F F = the distance between the proton and the electron is 29 2 2 1 9 2 31 2 (9 10 N m /C )(1.6 10 C) 5.08 m (9.11 10 kg)(9.8 m/s ) e ke r × == = × ASSESS The distance is almost fifty billion atomic diameters (or angstroms). This demonstrates that gravity is unimportant on the molecular scale.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap20 - ELECTRIC CHARGE, FORCE, AND FIELD 20 EXERCISES...

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