682_Physics ProblemsTechnical Physics

# 682_Physics ProblemsTechnical Physics - 22 Electric Fields...

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22 Electric Fields P23.58 From Figure A: d cos . . 30 0 15 0 °= cm, or d = ° 15 0 30 0 . cos . cm From Figure B: θ = F H G I K J sin . 1 50 0 d cm = ° F H G I K J sin . cos . . 1 15 0 30 0 20 3 cm 50.0 cm af F mg q = tan or Fm g q tan . 20 3 (1) From Figure C: FF q 23 0 0 cos . F kq q e = L N M M O Q P P ° 2 0300 30 0 2 2 . cos . m (2) Combining equations (1) and (2), 2 30 0 20 3 2 2 mg e . cos . tan . m L N M M O Q P P ° q mg k q q e 2 2 2 3 2 9 14 7 203 0 0 200 10 980 2 8 99 10 30 0 4 20 10 2 05 10 0 205 = ° ° = ×° ×⋅ ° = × = −− .t a n . cos . .. . t a n . .c o s . . m kg m s m Nm C C C C 2 22 2 ej e j µ Figure A Figure B Figure C FIG. P23.58 P23.59 Charge Q 2 resides on each block, which repel as point charges: F kQ Q L kL L e i == 2 bg . Solving for Q , QL k i e = 2 . *P23.60 If we place one more charge q at the 29th vertex, the total force on the central charge will add up to zero: F 28 charges + kqQ a e 2 away from vertex 29 0 = F 28 charges e k toward vertex 29 =
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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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