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# EXAM 1 - Platt David Quiz 1 Due 10:00 pm Inst Ken Shih This...

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Platt, David – Quiz 1 – Due: Sep 20 2005, 10:00 pm – Inst: Ken Shih 1 This print-out should have 27 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 dipole field pattern is shown in the figure. Consider various relationships between the electric potential at different points given in the figure. D J C P T + - Notice: Five potential relationships are given below. a) V J = V P > V C b) V J = V P = V C c) V J = V P < V C d) V D < V C < V T e) V D > V C > V T Which relations shown above are correct? 1. ( c ) and ( d ) only 2. ( b ) and ( e ) only 3. ( c ) only 4. ( a ) and ( d ) only 5. ( a ) and ( e ) only 6. ( a ) only 7. ( e ) only 8. ( b ) and ( d ) only correct 9. ( d ) only 10. ( c ) and ( e ) only Explanation: The electric potential due to one single point charge at a distance r from the charge is given by V = k q r . For a dipole system, the total potential at any place is the sum of potentials due to one positive point charge and one negative point charge (Superposition Principle). From symmetry considerations, it is easy to see that the electric field lines are perpen- dicular to a line which passes through the midpoint C and points B and D . No work needs to be done to move a positive test charge along the midplane because the force and the displacement are perpendicular to each other. V J = V C = V P , relation ( b ). Furthermore, moving along the direction of a electric field line ( i.e. , moving in the direc- tion from positive charge to negative charge along the electric field line) always lowers the electric potential, because the electric field will do positive work to a positive test charge in order to lower its electric potential energy. Therefore, V D < V J by considering the line going from J to D , and V P < V T by consider- ing the line going from T to P . V D < V C < V T , relation ( d ). The correct choices are ( b ) and ( d ) only. 002 (part 1 of 2) 10 points A point charge 4 q > 0 is placed at the center point O . There is a thick conducting spherical shell with inner radius R 2 and outer radius R 0 2 centered at O . The thickness of this shell is R 0 2 - R 2 . Another larger thin concentric spherical shell has radius R 3 . The thickness of this shell is negligible. The thick shell is charged with a charge 3 q and the large thin shell is charged with a charge 9 q .

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Platt, David – Quiz 1 – Due: Sep 20 2005, 10:00 pm – Inst: Ken Shih 2 A B O R 2 R 0 2 R 3 9 q 3 q 4 q t hi n s he l l t h i c k s h e ll The electric field at A, where OA = a > R 3 , is given by 1. E A = 4 q π ² 0 a 2 correct 2. E A = 15 4 q π ² 0 a 2 3. E A = 17 4 q π ² 0 a 2 4. E A = 19 4 q π ² 0 a 2 5. E A = 3 q π ² 0 a 2 6. E A = 9 2 q π ² 0 a 2 7. E A = 6 q π ² 0 a 2 8. E A = 5 q π ² 0 a 2 9. E A = 7 2 q π ² 0 a 2 Explanation: Let : q 1 = 4 q , q 2 = 3 q , and q 3 = 9 q . A B O R 2 R 0 2 R 3 q 3 q 2 q 1 t hi n s he l l t h i c k s h e ll Pick a Gaussian spherical surface centered at O, of radius a . The area of this surface is A = 4 πa 2 , so if the field is E , the flux through this surface is Φ = E A = E 4 πa 2 .
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