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

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Unformatted text preview: 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 1 2 3 4 5 x ( m )- 5- 4- 3- 2- 1 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 µ y a x a ¶ = arctan µ- 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 sin14 . 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 µ y b x b ¶ = arctan µ 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 sin146 . 31 ◦ = 1 . 53397 × 10- 9 N . 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 µ y c x c ¶ = arctan µ 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 ) sin36 . 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 ....
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This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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

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