Old Exam 1 Sol - Platt, David Oldquiz 1 Due: Sep 18 2005,...

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Unformatted text preview: Platt, David Oldquiz 1 Due: Sep 18 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A circular ring made of an insulating material is cut in half. One half is given a charge- q uniformly distributed along its arc. The other half is given a charge + q also uniformly distributed along its arc. The two halves are then rejoined with insulation at the junctions J , as shown. x y- q + q J J------------------ + + + + + + + + ++ + + + + + + + + q e If there is no change in the charge distri- butions, what is the direction of the net elec- trostatic force on an electron located at the center of the circle? 1. to the right 2. toward the bottom of the page correct 3. to the left 4. into the page 5. toward the top of the page Explanation: Pay attention to the symmetry in the con- figuration. The whole system is symmetrical about the diameter perpendicular to the line connecting two junctions. Also, we know the electrostatic force is along the direction connecting the two charges; repulsive if two charges are of the same sign and attractive if two charges are of the opposite sign. There- fore, the force on the electron is parallel to the symmetrical axis and points to the pos- itive charge; i.e. , toward the bottom of the page. 002 (part 1 of 1) 10 points From the electric field vector at a point, one can determine which of the following? I) the direction of the electrostatic force on a test charge of known sign at that point; II) the magnitude of the electrostatic force exerted per unit charge on a test charge at that point; III) the electrostatic charge at that point. 1. III only 2. I, II and III 3. II and III only 4. I and II only correct 5. I only Explanation: The definition of the electrostatic force is ~ E = ~ F q . In another way, ~ F = q ~ E. This means ~ F is in the same direction of, or opposite di- rection to ~ E , depending on the sign of the charge. And if we only consider the magni- tude F = qE for a unit charge, the force on it is just F q = E . This is statement II. 003 (part 1 of 2) 10 points Consider a square with side a . Four charges + q , + q ,- q , and + q are placed at the corners A , B , C , and D , respectively. + +- + D C A B a The magnitude of the electric field at D due to the charges at A , B , and C is given by Platt, David Oldquiz 1 Due: Sep 18 2005, 4:00 am Inst: Ken Shih 2 1. k ~ E k = 7 2 k q a 2 2. k ~ E k = 2 k q a 2 3. k ~ E k = 2 k q a 2 4. k ~ E k = 3 k q a 2 5. k ~ E k = k q a 2 6. k ~ E k = 5 2 k q a 2 7. k ~ E k = 3 2 k q a 2 correct 8. k ~ E k = 9 4 k q a 2 Explanation: The magnitudes of the electric fields at D due to A and C are E A = E C = k q a 2 since they are at a distance a , whereas E B = k q ( a 2) 2 = k q 2 a 2 since B is at a distance 2 a ....
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Old Exam 1 Sol - Platt, David Oldquiz 1 Due: Sep 18 2005,...

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