Exam 1-solutions - Version 085 Exam 1 markert(55325 This...

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

View Full Document Right Arrow Icon
Version 085 – Exam 1 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two charged particles of equal magnitude ( Q and Q ) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge + q is placed at a third corner. Q + q Q What is the direction of the force on the test charge due to the two other charges? 1. 2. 3. 4. 5. 6. 7. 8. correct Explanation: The force between charges of the same sign is repulsive and the force between charges with opposite signs is attractive. Q + q Q The resultant force is the sum of the two vectors in the figure. 002 10.0points Consider a charge q 1 on a metallic ball with radius R 1 at the center, inside of a concentric conducting shell with total charge q 2 of inner radius R 2 and outer radius R 3 . O q 2 q 1 A R 1 R 2 R 3 Find the electric field at A , where OA = a . 1. E A = k q 1 + q 2 R 2 3 2. E A = k q 1 + q 2 a 2 correct 3. E A = k q 1 2 a 2 4. E A = 0 5. E A = k q 1 q 2 R 2 3 6. E A = k q 1 R 2 3 7. E A = k q 1 q 2 a 2 8. E A = k q 1 q 2 2 a 2 9. E A = k q 1 a 2 10. E A = k q 1 + q 2 2 a 2
Image of page 1

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

View Full Document Right Arrow Icon
Version 085 – Exam 1 – markert – (55325) 2 Explanation: Consider a spherical Gaussian surface through A . A is outside of the entire charge distribution a distance a from the center, so the enclosed charge Q encl = q 1 + q 2 can be treated as a point charge, and the electric field is E A = k q 1 + q 2 a 2 . 003 10.0points 1) Two uncharged metal balls, X and Y , each stand on a glass rod and are touching. Y X 2) A third ball carrying a negative charge, is brought near the first two. Y X 3) While the positions of these balls are fixed, ball X is connected to ground. Y X 4) Then the ground wire is disconnected. Y X 5) While X and Y remain in touch, the ball carring the negative charge is removed. Y X 6) Then ball X and Y are separated. Y X After these procedures, what are the signs of the charge q X on X and q Y on Y ? 1. q X is positive and q Y is negative 2. q X is positive and q Y is positive correct 3. q X is neutral and q Y is negative 4. q X is positive and q Y is neutral 5. q X is neutral and q Y is positive 6. q X is neutral and q Y is neutral 7. q X is negative and q Y is positive 8. q X is negative and q Y is negative 9. q X is negative and q Y is neutral Explanation: When the ball with negative charge is brought nearby, the free charges inside X and Y rearrange themselves. The positive charges are attracted and go to the left ( i.e. , move to Y ), leaving negative charges on the right hand side of the system XY , i.e., in X . When we ground X , electrons flow from the ground to X (making it neutral), whereas the positive charges in Y are still held enthralled by the negative charge on the third ball. We break the ground.
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