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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern