20.1ELECTRIC CHARGE, FORCE, AND FIELD 20EXERCISESSection 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 approximately272812(65 kg)/(1.6710kg)210 ,−×<×0,ewhich 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 aboutor several coulombs (huge by ordinary standards). 9protonelectron||1qq−−=28919210101.610C3.2 C,−−± ××××= ±15. INTERPRETThis problem deals with quantity of charge in a typical lightning flash. We want to express the quantity in terms of elementary charge e. DEVELOPSince the magnitude of elementary charge eisthe number of electrons involved is given by191.610C,e−=×/ .Q eEVALUATESubstituting the values given in the problem statement, we find the number to be 201925 C/1.1.610CNQ e−===××5610ASSESSSince 1 coulomb is aboutelementary charges, our result has the right order of magnitude. 186.2510×16.(a)The proton’s charge is2213331corresponding to a combination of uudquarks; (b)for neutrons,,eeee=+−0=211333−−corresponds to udd. (See Chapter 39, or Chapter 45 in the extended version of the text.) Section 20.2 Coulomb’s Law 17. INTERPRETIn this problem we are asked to compare the gravitational and electrical forces between a proton and an electron. DEVELOPThe gravitational force and the electrostatic force between a proton and an electron separated by a distance rare, respectively, grav2peGm mFr=and 2elec2keFr=EVALUATEThe ratio of the two forces is 22922192elec2112227grav39(910Nm /C)(1.610C)(6.6710Nm /kg)(1.6710kg)(9.1110kg)2.310peFkerFGm mr−−−⎛⎞⎛⎞×⋅×==⎜⎟⎜⎟⎜⎟×⋅××⎝⎠⎝⎠≈×31−Note that the spatial dependence of both forces is the same, and cancels out. ASSESSAt all distances (for which the particles can be regarded as classical point charges), the Coulomb force is about10times stronger than the gravitational force. 40
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