683_Physics ProblemsTechnical Physics

683_Physics ProblemsTechnical Physics - 23 Chapter 23...

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Chapter 23 23 P23.62 At equilibrium, the distance between the charges is r = × 20100 100 347 10 2 .s i n . . m m bg Now consider the forces on the sphere with charge + q , and use F y = 0: F y = Tm g cos . 10 0 °= , or T mg = ° cos . 10 0 (1) F x = 0 : FF F T net =−= ° 21 10 0 sin . (2) F net is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1). F mg mg net 2 kg m s N = ° ° = × ° = × −− sin . cos . tan . . . tan . . 10 0 10 0 10 0 2 00 10 9 80 10 0 3 46 10 33 ej e j F net is the resultant of two forces, F 1 and F 2 . F 1 is the attractive force on + q exerted by q , and F 2 is the force exerted on + q by the external electric field. r L +q –q θθ FIG. P23.62 F net =− or FF F =+ net F 1 9 88 3 2 2 899 10 5 00 10 5 00 10 347 10 187 10 ×× × . .. . . Nm C C C m N 22 Thus, net yields F 2 32 2 3 46 10 1 87 10 2 21 10 ... N N N and Fq E 2 = , or E F q == × × = 2 2 8 5 221 10 443 10 443 . . N 5.00 10 C NC kNC . P23.63 Qd R d R R R Q dF d R Rd R F y y = = = = = = F H G I K J = F H G G I K J J × zz −° ° ° λλ θ λ θλ µλ µ π A A 0 90 0 90 0 0 90 0 90 0 00 0 0 2 0 0 2 2 9 112 12 0 2 0 600 12 0 10 0 1 4 300 1 4 300 10 cos sin . . . cos .c o s . . . . . . af b g m C s o C m C C 0 66 2 90 0 90 0 3 2 2 2 2 10 0 10 0600 899300 10 1 2 1 2 2 0450 1 2 1 4 20 7 0 7 C m m N N N Downward. e j . . cos . cos i n . . . × F H G I K J F H G I K J = °
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