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Unformatted text preview: Kapoor (mk9499) – homework10 – Turner – (60230) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A dipole field pattern is shown in the figure. Consider various relationships between the electric potential at different points given in the figure. K R G D L + Notice: Five potential relationships are given below. a) V R = V D > V G b) V R = V D = V G c) V R = V D < V G d) V K < V G < V L e) V K > V G > V L Which relations shown above are correct? 1. ( c ) and ( e ) only 2. ( e ) only 3. ( c ) and ( d ) only 4. ( b ) and ( d ) only correct 5. ( d ) only 6. ( a ) and ( d ) only 7. ( c ) only 8. ( a ) only 9. ( b ) and ( e ) only 10. ( a ) 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 G and points R 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 R = V G = V D , 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 K < V R by considering the line going from R to K , and V D < V L by consid ering the line going from L to D . V K < V G < V L , relation ( d ). The correct choices are ( b ) and ( d ) only. 002 (part 1 of 4) 10.0 points Two charges are located in the ( x,y ) plane as shown in the figure below. The fields pro duced by these charges are observed at the origin, p = (0 , 0). The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Kapoor (mk9499) – homework10 – Turner – (60230) 2 x y + Q + Q p b a a Use Coulomb’s law to find the xcomponent of the electric field at p . 1. E x = 0 correct 2. E x = 4 k e Qa ( a 2 + b 2 ) 3 / 2 3. E x = 2 k e Q a 2 + b 2 4. E x = 2 k e Qa ( a 2 + b 2 ) 3 / 2 5. E x = 2 k e Q a 2 + b 2 6. E x = k e Qa ( a 2 + b 2 ) 3 / 2 7. E x = 4 k e Qa ( a 2 + b 2 ) 3 / 2 8. E x = k e Qa ( a 2 + b 2 ) 3 / 2 9. E x = 2 k e Qa ( a 2 + b 2 ) 3 / 2 Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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