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Unformatted text preview: song (shs546) – Practice HW 2 Solutions – weathers – (22202) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose that 1 . 7 g of hydrogen (H 2 ) is sep arated into electrons and protons, and that the protons are placed at the Earth’s North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth? The radius of the Earth is 6 . 37 × 10 6 m. Avogadro’s number is 6 . 02214 × 10 23 and the molar mass of the Hatom is 1 . 00782 g / mole. Correct answer: 1 . 46674 × 10 6 N. Explanation: Let : m = 1 . 7 g = 0 . 0017 kg , R E = 6 . 37 × 10 6 m , N A = 6 . 02214 × 10 23 mol − 1 , q e = 1 . 60218 × 10 − 19 C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . The mass m is proportional to n , the num ber of hydrogen atoms. The number of moles of hydrogen can be expressed two ways: m M H = n N A n = N A M H m = parenleftbigg 6 . 02214 × 10 23 mol − 1 1 . 00782 g / mol parenrightbigg (1 . 7 g) = 1 . 01581 × 10 24 atoms . Since each of these atoms is split into a proton and an electron, there will be n protons at the North Pole and n electrons at the South Pole, and the force is F = k e parenleftbigg nq e 2 R E parenrightbigg 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × bracketleftbigg 1 . 01581 × 10 24 2 (6 . 37 × 10 6 m) bracketrightbigg 2 × ( 1 . 60218 × 10 − 19 C ) 2 = 1 . 46674 × 10 6 N . 002 10.0 points Two charges are located in the ( x,y ) plane as shown in the figure below. The fields pro duced by these charges are observed at a point p with coordinates (0 , 0). 7 . 3 C − 8 . 5 C p 1 . 7 m 1 . 5 m 2 . 8 m 2 . 2 m Use Coulomb’s law to find the xcomponent of the electric field at p . The Coulomb con stant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: − 1 . 41292 × 10 10 N / C. Explanation: Let : ( x p ,y p ) = (0 , 0) , ( x 1 ,y 1 ) = (2 . 8 m , − 1 . 7 m) , ( x 2 ,y 2 ) = ( − 2 . 2 m , − 1 . 5 m) , r = radicalbig x 2 + y 2 , q 1 = 7 . 3 C , q 2 = − 8 . 5 C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 q 2 p y 1 y 2 x 1 x 2 The figure below shows the electric field vectors. song (shs546) – Practice HW 2 Solutions – weathers – (22202) 2 θ 1 θ 2 E 1 E 2 7 . 3 C − 8 . 5 C where θ 1 = 180 ◦ − tan − 1 vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 7 m 2 . 8 m vextendsingle vextendsingle vextendsingle vextendsingle = 148 . 736 ◦ , θ 2 = 180 ◦ + tan − 1 vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 5 m − 2 . 2 m vextendsingle vextendsingle vextendsingle vextendsingle = 214 . 287 ◦ ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler

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