TEST01-solutions - Version 004 TEST01 tsoi (57210) 1 This...

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Unformatted text preview: Version 004 TEST01 tsoi (57210) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Three small spheres carry equal amounts of electric charge. They are equally spaced and lie along the same line. + + + What is the direction of the net electric force on each charge due to the other charge? 1. + + + correct 2. + + + 3. + + + 4. + + + 5. + + + 6. + + + 7. + + + 8. + + + 9. + + + 10. + + + Explanation: Since like charges repel and unlike charge attract, + + + 002 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 1 r 1 b V 2 vector E 2 I 2 2 r 2 b If 2 = 1 ,r 2 = 2 r 1 , 2 = 3 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 2 3 2. R 2 R 1 = 2 3. R 2 R 1 = 1 2 4. R 2 R 1 = 4 5. R 2 R 1 = 1 4 6. R 2 R 1 = 1 3 Version 004 TEST01 tsoi (57210) 2 7. R 2 R 1 = 3 2 8. R 2 R 1 = 3 4 correct 9. R 2 R 1 = 3 10. R 2 R 1 = 4 3 Explanation: The relation between resistance and resis- tivity is given by R = A = r 2 . Thus the ratio of the resistances is R 2 R 1 = 2 r 2 2 r 2 1 1 = 2 1 r 2 1 r 2 2 = 3 1 1 r 2 1 (2 r 1 ) 2 = 3 4 . 003 10.0 points A cubic box of side a , oriented as shown, con- tains an unknown charge. The vertically di- rected electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. a E 2 E How much charge Q is inside the box? 1. Q encl = 1 2 E a 2 2. Q encl = 2 E a 2 3. Q encl = 6 E a 2 4. Q encl = E a 2 5. Q encl = 3 E a 2 6. Q encl = 2 E a 2 7. Q encl = E a 2 correct 8. Q encl = 3 E a 2 9. insufficient information 10. Q encl = 0 Explanation: Electric flux through a surface S is, by con- vention, positive for electric field lines going out of the surface S and negative for lines going in. Here the surface is a cube and no flux goes through the vertical sides. The top receives top = E a 2 (inward is negative) and the bottom bottom = 2 E a 2 . The total electric flux is E = E a 2 + 2 E a 2 = E a 2 . Using Gausss Law, the charge inside the box is Q encl = E = E a 2 . 004 10.0 points In this problem assume that the batteries have zero internal resistance. The currents are flowing in the direction indicated by the arrows. Version 004 TEST01 tsoi (57210) 3 15 . 9 21 . 5 8 . 2 V 17 . 3 V I 1 I 2 I 3 Find the current through the 15 . 9 resistor and the 8 . 2 V battery at the top of the circuit....
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TEST01-solutions - Version 004 TEST01 tsoi (57210) 1 This...

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