666_Physics ProblemsTechnical Physics

666_Physics ProblemsTechnical Physics - 6 Electric Fields...

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6 Electric Fields P23.6 We find the equal-magnitude charges on both spheres: Fk qq r k q r ee == 12 2 2 2 so qr F k e × ×⋅ 100 100 10 105 10 4 3 . . . m N 8.99 10 N m C C 92 2 af . The number of electron transferred is then N e xfer C C electrons = × × −− 160 10 659 10 3 19 15 . . .. The whole number of electrons in each sphere is Ne e tot g 107.87 g mol atoms mol atom = F H G I K J ×= × 10 0 6 02 10 47 2 62 10 23 24 . ej e j . The fraction transferred is then f N N × × F H G I K J = xfer tot 262 10 251 10 251 15 24 9 . . . . charges in every billion. P23.7 r e 1 2 96 6 2 8 99 10 7 00 10 2 00 10 0500 0503 × × = . . . Nm C C C m N 22 e j e j a f r e 2 2 6 2 8 99 10 7 00 10 4 00 10 101 × × = . . . C C m N e j a f F F x y + ° = ° = =−= ° 0 503 60 0 1 01 60 0 0 755 0 503 60 0 1 01 60 0 0 436 0 755 0 436 0 872 .c o s . . c o s . . .s i n . . s i n . . . ± . ± . N N N N N at an angle of 330 Fij FIG. P23.7 P23.8 r e × × × = 2 91 9 2 23 2 6 2 8 99 10 1 60 10 6 02 10 2 6 37 10 514 . . C m kN e j e j P23.9 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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