{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz2_sol - Physics 2B Quiz 2 Name Summer Session I 2009...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 2B Quiz 2 Summer Session I 2009 Version 0 Name: PID: Quiz ID: READ FIRST: Make sure to write/fill in your scantron form your name, quiz ID and the test version. e 1 . 6 × 10 - 19 C , k 9 . 0 × 10 9 N m 2 / C 2 , 0 8 . 9 × 10 - 12 C 2 / N m 2 , Z dx x - a = ln | x - a | Unless otherwise specified, assume potential is zero at infinity. 1. An electron move in the - ˆ x direction 0 . 5m through an electric field ~ E = 2 . x N / C. What is the change in potential energy? Solution: The potential difference is: Δ V = - ~ E · ~ l = 1 . 25V. An electron carries charge - e , So the change in potential energy is ΔU = q Δ V = - 1 . 25 eV. A. -1.25 eV B. 1.25 eV C. 1.25 J D. -1.25 J 2. At each of the four corners of a square is a 2 nC charge. The side of the square is 1.41 m. What is electric potential at the center of the square? Solution: The center is the same distance away from each charges: 1 . 41 / 2 1 (m). Calculate the potential using principle of superposition: φ = 4 kq d 4 × 9 × 10 9 J · m C 2 · 2 × 10 - 9 C 1 m = 72V A. 72 V B. 36 V C. 18 V D. 144 V 3. How much energy is needed to assemble four 2 nC charges to form a square of side 1.41 cm? Solution: Each charge see three other charges: two at distance 1.41 cm, one at distance 2 cm. So the total energy is just four times the energy to move the last charge in placewith the factor of 1 / 2 to take into account double-counting.
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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