Practice HW 2 Solutions

Practice HW 2 Solutions - lim(kl9356 – Practice HW 2...

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Unformatted text preview: lim (kl9356) – Practice HW 2 Solutions – Weathers – (22202) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose that 1 . 4 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. 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 H-atom is 1 . 00782 g / mole. What is the resulting compressional force on the Earth? Correct answer: 9 . 94743 × 10 5 N. Explanation: Let : m = 1 . 4 g = 0 . 0014 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 total mass m di- vided by the molar weight of hydrogen M H is the number of moles of atomic hydrogen. Sim- ilarly, the total number of hydrogen atoms n divided by Avogadro’s number N A also gives the number of moles of hydrogen. Thus, m and n are related, 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 . 4 g) = 8 . 36553 × 10 23 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 8 . 36553 × 10 23 2 (6 . 37 × 10 6 m) bracketrightbigg 2 × ( 1 . 60218 × 10 − 19 C ) 2 = 9 . 94743 × 10 5 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). The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 7 . 6 C − 7 . 8 C p 1 . 6 m 1 . 7 m 2 . 4 m 2 . 7 m Use Coulomb’s law to find the x-component of the electric field at p . Correct answer: − 1 . 26584 × 10 10 N / C. Explanation: Let : ( x p ,y p ) = (0 , 0) , ( x 1 ,y 1 ) = (2 . 4 m , − 1 . 6 m) , ( x 2 ,y 2 ) = ( − 2 . 7 m , − 1 . 7 m) , r = radicalbig x 2 + y 2 , q 1 = 7 . 6 C , q 2 = − 7 . 8 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 lim (kl9356) – Practice HW 2 Solutions – Weathers – (22202) 2 The figure below shows the electric field vectors. θ 1 θ 2 E 1 E 2 7 . 6 C − 7 . 8 C where θ 1 = 180 ◦ − tan − 1 vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 6 m 2 . 4 m vextendsingle vextendsingle vextendsingle vextendsingle = 146 . 31 ◦ , θ 2 = 180 ◦ + tan − 1 vextendsingle vextendsingle vextendsingle vextendsingle − 1 . 7 m − 2 . 7 m vextendsingle vextendsingle vextendsingle vextendsingle = 212 . 196 ◦ ....
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 2 Solutions - lim(kl9356 – Practice HW 2...

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