HW2Sol - Platt David Homework 2 Due Sep 9 2005 4:00 am Inst...

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Platt, David – Homework 2 – Due: Sep 9 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 0 points Four charges - 2 × 10 - 9 C at (0 m,0 m), - 1 × 10 - 9 C at ( - 4 m, - 1 m), 2 × 10 - 9 C at ( - 3 m,2 m), and 5 × 10 - 9 C at (4 m,3 m), are arranged in the ( x, y ) plane (as shown in the figure below, where the scale is in meters). y ( m ) - 5 - 3 - 1 0 1 2 3 4 5 x ( m ) - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 - 2 nC - 1 nC 2 nC 5 nC Find the magnitude of the resulting force on the - 2 nC charge at the origin [coordinates, (0 m , 0 m)]. Correct answer: 4 . 25969 × 10 - 9 N. Explanation: Let : q o = - 2 × 10 - 9 C , ( x o , y o ) = (0 m , 0 m) , q a = - 1 × 10 - 9 C , ( x a , y a ) = ( - 4 m , - 1 m) , q b = 2 × 10 - 9 C , ( x b , y b ) = ( - 3 m , 2 m) , q c = 5 × 10 - 9 C , and ( x c , y c ) = (4 m , 3 m) . Coulomb’s Law for q o and q a is F ao = k e q o q a ( x a - x o ) 2 + ( y a - y o ) 2 = ( 8 . 98755 × 10 9 N · C 2 / m 2 ) × ( - 2 × 10 - 9 C ) ( - 1 × 10 - 9 C ) ( - 4 m) 2 + ( - 1 m) 2 = 1 . 05736 × 10 - 9 N . θ a = arctan y a x a = arctan - 1 m - 4 m = 14 . 0362 . F a x = F a cos θ a = fl fl 1 . 05736 × 10 - 9 N fl fl cos 14 . 0362 = 1 . 02579 × 10 - 9 N . F a y = F a sin θ a = fl fl 1 . 05736 × 10 - 9 N fl fl sin 14 . 0362 = 2 . 56447 × 10 - 10 N . Coulomb’s Law for q o and q b is F bo = k e q o q b ( x b - x o ) 2 + ( y b - y o ) 2 = ( 8 . 98755 × 10 9 N · C 2 / m 2 ) × ( - 2 × 10 - 9 C ) ( 2 × 10 - 9 C ) ( - 3 m) 2 + (2 m) 2 = - 2 . 7654 × 10 - 9 N . θ b = arctan y b x b = arctan 2 m - 3 m = 146 . 31 . F b x = F b cos θ b = fl fl - 2 . 7654 × 10 - 9 N fl fl cos 146 . 31 = - 2 . 30095 × 10 - 9 N . F b y = F b sin θ b = fl fl - 2 . 7654 × 10 - 9 N fl fl sin 146 . 31 = 1 . 53397 × 10 - 9 N .
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Platt, David – Homework 2 – Due: Sep 9 2005, 4:00 am – Inst: Ken Shih 2 Coulomb’s Law for q o and q c is F co = k e q o q c p ( x c - x o ) 2 + ( y c - y o ) 2 = ( 8 . 98755 × 10 9 N · C 2 / m 2 ) × ( - 2 × 10 - 9 C ) ( 5 × 10 - 9 C ) p (4 m) 2 + (3 m) 2 = - 3 . 59502 × 10 - 9 N . θ c = arctan y c x c = arctan 3 m 4 m = 36 . 8699 . F c x = F c cos θ c = ( - 3 . 59502 × 10 - 9 N ) cos 36 . 8699 = 2 . 87602 × 10 - 9 N . F c y = F c sin θ c = ( - 3 . 59502 × 10 - 9 N ) sin 36 . 8699 = 2 . 15701 × 10 - 9 N . The magnitude of the resultant force is F = £ F 2 x + F 2 y / 1 / 2 = h ( F a x + F b x + F c x ) 2 + ( F a y + F b y + F c y ) 2 i 1 / 2 = £( 1 . 02579 × 10 - 9 N + - 2 . 30095 × 10 - 9 N 2 . 87602 × 10 - 9 N ) 2 + ( 2 . 56447 × 10 - 10 N 1 . 53397 × 10 - 9 N +2 . 15701 × 10 - 9 N ) 2 i 1 / 2 = 4 . 25969 × 10 - 9 N . 002 (part 2 of 2) 0 points Select the figure showing the direction of the resultant force on the - 2 nC charge at the origin. 1. y θ = 150 . 564 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 2. y θ = 280 . 523 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 3. y θ = 49 . 5175 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 4. y θ = 17 . 602 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 5. None of these figures is correct.
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Platt, David – Homework 2 – Due: Sep 9 2005, 4:00 am – Inst: Ken Shih 3 6. y θ = 266 . 282 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 7. y θ = 128 . 45 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 8. y θ = 67 . 9253 - 5 - 3 - 10 1 2 3 4 5 x - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 cor- rect Explanation: The direction θ as measured in a counter- clockwise direction from the positive x axis is θ = arctan F y F x = arctan 3 . 94743 × 10 - 9 N 1 . 60085 × 10 - 9 N = 67 . 9253 .
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