solution_pdf_hw1

# solution_pdf_hw1 - royal(jhr696 oldhomework 01 Turner(56705...

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royal (jhr696) – oldhomework 01 – Turner – (56705) 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 The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 0 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1. E A E B = 1 1 . 2. E A E B = 2 1 . 3. E A E B = 8 1 . 4. E A E B = 4 1 . correct 5. E A E B = 1 2 . Explanation: Let : r B = 2 r A . The electric field strength E 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 002 10.0 points Suppose that 1 . 2 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 H-atom is 1 . 00782 g / mole. Correct answer: 7 . 30831 × 10 5 N. Explanation: Let : m = 1 . 2 g = 0 . 0012 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 . 2 g) = 7 . 17046 × 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 n q e 2 R E parenrightbigg 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × bracketleftbigg 7 . 17046 × 10 23 2 (6 . 37 × 10 6 m) bracketrightbigg 2 × ( 1 . 60218 × 10 19 C ) 2 = 7 . 30831 × 10 5 N . 003 10.0 points Three identical point charges hang from three strings, as shown.

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royal (jhr696) – oldhomework 01 – Turner – (56705) 2 45 45 F g 39.0 cm 39.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 2 . 57696 × 10 6 C. Explanation: Let : m = 0 . 10 kg , L = 39 . 0 cm , θ = 45 , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r = 2 L sin θ = 2 L sin 45 = 2 L 2 2 = L 2 F T,x = F T sin θ F T,y = F T cos θ Each sphere is in equilibrium horizontally F electric F T,x = 0 F electric F T sin θ = 0 and vertically F T,y F g = 0 F T cos θ F g = 0 F T = F g cos θ . From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg sin θ F electric = F g tan θ = F g (tan 45 ) = F g . For either of the end charges, F electric = k e q 2 r 2 + k e q 2 ( r 2 ) 2 = k e q 2 r 2 + 4 k e q 2 r 2 = 5 k e q 2 r 2 5 k e q 2 r 2 = m g . Thus | q | = radicalBigg r 2 m g 5 k e = radicalBigg ( L 2) 2 m g 5 k e = L · radicalbigg 2 m g 5 k e = (39 cm) parenleftbigg 1 m 100 cm parenrightbigg × radicalBigg 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 2 . 57696 × 10 6 C .
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