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# solution1_pdf - taherisefat(mt23852 hw 1 Opyrchal(121012...

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taherisefat (mt23852) – hw 1 – Opyrchal – (121012) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge of +1 coulomb is place at the 0- cm mark of a meter stick. A charge of +4 coulombs is placed at the 100-cm mark of the same meter stick. Where should a proton be placed on the meter stick so that the net force on it due to the two charges is 0? Correct answer: 33 . 3333 cm. Explanation: The forces are both repulsive and point in opposite directions. The force due to the +1 charge is F 1 = k (+1) (+1) r 2 1 = k r 2 1 and due to the +4 charge, F 4 = k (+4) (+1) r 2 4 = 4 k r 2 4 . In order for these forces to be equal, k r 2 1 = 4 k r 2 4 r 2 4 = 4 r 2 1 = (2 r 1 ) 2 r 4 = 2 r 1 The ratio of the two distances is 1:2, so r 1 = 1 3 (100 cm) = 33 . 3333 cm . 002 (part 1 of 2) 10.0 points Three charges are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). 1 nC 8 nC 6 nC y ( m ) 0 1 2 3 4 5 6 7 8 9 10 x ( m ) 0 1 2 3 4 5 6 7 8 9 10 What is the magnitude of the resulting force on the 1 nC charge at the origin? Correct answer: 1 . 10766 nN. Explanation: Let : q o = 1 × 10 9 C , ( x o , y o ) = (0 m , 0 m) , q a = 8 × 10 9 C , ( x a , y a ) = (10 m , 0 m) , q b = 6 × 10 9 C , and ( x b , y b ) = (0 m , 8 m) . Coulomb’s Law for q o and q a is bardbl vector F oa bardbl = k e q o q a radicalbig ( x a x o ) 2 + ( y a y o ) 2 = k e q o q a radicalbig x 2 oa + y 2 oa = 8 . 98755 × 10 9 N C 2 / m 2 × (1 × 10 9 C) ( 8 × 10 9 C) radicalbig (10 m) 2 + (0 m) 2 = 7 . 19004 × 10 10 N . F oa x = k e q o q a r 2 oa cos θ oa = k e q o q a r 2 oa x oa r oa = (8 . 98755 × 10 9 N C 2 / m 2 ) × (1 × 10 9 C) ( 8 × 10 9 C) (10 m) 2 × (10 m) (10 m) = 7 . 19004 × 10 10 N .

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taherisefat (mt23852) – hw 1 – Opyrchal – (121012) 2 F oa y = k e q o q a r 2 oa sin θ oa = k e q o q oa r 2 oa y oa r oa = (8 . 98755 × 10 9 N C 2 / m 2 ) × (1 × 10 9 C) ( 8 × 10 9 C) (10 m) 2 × (0 m) (10 m) = 0 N . Coulomb’s Law for q o and q b is bardbl vector F ob bardbl = k e q o q b radicalbig ( x b x o ) 2 + ( y b y o ) 2 = k e q o q b radicalBig x 2 ob + y 2 ob = 8 . 98755 × 10 9 N C 2 / m 2 × (1 × 10 9 C) (6 × 10 9 C) radicalbig (0 m) 2 + (8 m) 2 = 8 . 42583 × 10 10 N .
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solution1_pdf - taherisefat(mt23852 hw 1 Opyrchal(121012...

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