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Physics 201 Homework Solutions 01 CHAPTER 21: Electric Charge & Field 21[6, 8, 13, 17, 27, 42, 56*, 63, 100, 88] *Problems are Optional 1 Problems 21–5 Coulomb’s Law3691mC10C, 1C10C, 1nC10C.µ−===--6. Since the magnitude of the force is inversely proportional to the square of the separation distance, 21Fr∝, if the distance is multiplied by a factor of 1/8, the force will be multiplied by a factor of 64. ()2064643.210N2.0 NFF−==×=8. Use the charge per electron and the mass per electron. ()()61414193114161 electron4610C2.871102.910 electrons1.60210C9.10910kg2.87110 e2.610kg1 e−−−−−−−×=×≈×−×××=×13. The forces on each charge lie along a line connecting the charges. Let the variable drepresent the length of a side of the triangle. Since the triangle is equilateral, each angle is 60o. First calculate the magnitude of each individual force. ()()()()()()()()66922121222669221313227.010C8.010C8.98810 N mC1.20m0.3495N7.010C6.010C8.98810 N mC1.20m0.2622 NQ QFkdQ QFkd−−−−××==×⋅=××==×⋅=()()()()66922232332228.010C6.010C8.98810 N mC0.2996 N1.20mQ QFkFd−−××==×⋅==Now calculate the net force on each charge and the direction of that net force, using components. ()()()()oo211213oo1112131122111111210.3495N cos600.2622 N cos604.36510N0.3495N sin600.2622 N sin605.29710N5.29710N0.53N tantan2654.36510NxxxyyyyxyxFFFFFFFFFFFθ−−−−−−= +=−+=−×= +=−−=−×−×=+====°−×1Q2Q3Q12Fd13Fdd32F31F21F23F
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