WHomework set 2_solns_2011

WHomework set 2_solns_2011 - Webassign Homework set # 2...

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Webassign Homework set # 2 Problems: 15.15, 16, 17, 22 Solutions: 15.15 For the object to “float” it is necessary that the electrical force support the weight, or     6 3 2 24 10 C 610 N C or 1.5 10 kg 9.8 m s qE qE mg m g  15.16 (a) Taking to the right as positive, the resultant electric field at point P is given by  132 222 26 6 6 9 2 N m 6.00 10 C 2.00 10 C 1.50 10 C 8.99 10 C 0.020 0 m 0.030 0 m 0.010 0 m R eee EE E E kq kq kq rrr        This gives 7 2.00 10 N C R E  or 7 2.00 10 N C to the right R E (b)     67 40.0 N R q FE  or 40.0 N to the left F
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15.17 We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration. The contribution due to the positive charge at 3 000 m altitude is    2 95 2 22 40.0 C Nm 8.99 10 3.60 10 N C downward C 1000 m e q Ek r   
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This note was uploaded on 05/12/2011 for the course PHYS 1112 taught by Professor Seaton during the Spring '08 term at University of Georgia Athens.

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WHomework set 2_solns_2011 - Webassign Homework set # 2...

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