14-3 - brown(twb493 – Ch14-h3 – chiu –(56565 1 This...

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Unformatted text preview: brown (twb493) – Ch14-h3 – chiu – (56565) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of these statements about a dipole are correct? 1 A dipole consists of two particles whose charges are equal in magnitude but oppo- site in sign 2 The net electric field due to a dipole is zero, since the contribution of the negative charge cancels out the contribution of the positive charge 3 At a distance d from a dipole, where d ≫ s (where s is the separation of the charges), the magnitude of the electric field due to the dipole is proportional to 1 d 3 4 At a distance d from a dipole, where d ≫ s (where s is the separation of the charges), the magnitude of the electric field due to the dipole is proportional to 1 d 2 5 When placed in a constant electric field, a dipole does not interact at all, since the force on the positive charge is in an opposite direction to the force on the negative charge Your answer should be a list of numbers corresponding to the correct statements, with the numbers separated by commas Correct answer: 1,3. Explanation: Let’s examine each of these claims individ- ually: 1 A dipole is defined to be a distribution of two equal and opposite charges. 2 Although the charges in a dipole have equal magnitude, they are in different locations For this reason, a typical point will be a different distance away from each of the two dipole charges, and the vector sum of the two fields will be nonzero. 3 We can derive this relationship using Coulomb’s law. Let the charge on the dipole be q , and their separation distance be a . For simplicity, consider a point on the line connecting the two charges, a distance x from the positive charge. Then, the net field lies on the x-axis, and has magnitude: E = k q r 2 − k q ( r + a ) 2 = kq parenleftbigg ( r + a ) 2 − r 2 ( r ( r + a )) 2 parenrightbigg = kq parenleftbigg r 2 + 2 ra + a 2 − r 2 ( r ( r + a )) 2 parenrightbigg = kq parenleftbigg 2 ra + a 2 ( r ( r + a )) 2 parenrightbigg So, when r is much larger than a , we have E ≈ kq r 3 . 4 We do not expect the field to fall off at the same rate for a dipole as it would for a single charge, since the field from the negative charge has a cancelling effect against the positive charge. 5 While the forces on the charges on a dipole are equal in magnitude and in opposite di- rections, they are applied in different loca- tions, which will create a net torque on the dipole, even if the net force is zero. 002 (part 1 of 2) 10.0 points A charge of 1 nC (1 nC = 1 × 10 − 9 C) and a dipole with charges + q and − q separated by . 3 mm contribute to a net field at location A that is zero, as shown in the following figure. brown (twb493) – Ch14-h3 – chiu – (56565) 2 b A 10 cm 18 cm + +1 nC Which end of the dipole is positively charged?...
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

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14-3 - brown(twb493 – Ch14-h3 – chiu –(56565 1 This...

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