This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 081 midterm 01 chiu (56565) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Two dipoles are oriented as shown in the diagram below x A r r Each dipole consists of two charges + q and q , held apart by a rod of length s , and the center of each dipole is a distance r from location A . If q = 5 nC, s = 2 mm, and r = 10 cm, what is the electric field ( E y ) at location A ? Hint: Draw a diagram and show the direc tion of each dipoles contribution to the elec tric field on the diagram (you do not have to turn in the diagram). The net electric field can be written in the form: vector E = ( E x , E y , E z ) where it is understood that E z = 0 by the planar nature of the problem. What is E y ? 1. 393.586 2. 185.185 3. 444.444 4. 421.875 5. 202.855 6. 592.593 7. 472.303 8. 270.0 9. 944.606 10. 243.426 Correct answer: 270 N / C. Explanation: Now, we have to be more careful about determining the electric field. First, we remember that the electric field due to a point charge is vector F = k q r 2 r = k q r 3 ( x, y, z ) So, lets consider the net field due to the left dipole first. First, note that the electric field due to both point charges has a positive y component since the electric field points away from positive charges and toward negative charges. Then, we have: E y,left = k q s 2 ( r 2 + ( s 2 ) 2 ) 3 / 2 + k q s 2 ( r 2 + ( s 2 ) 2 ) 3 / 2 = k qs ( r 2 + ( s 2 ) 2 ) 3 / 2 Finally, we can neglect the factor of parenleftBig s 2 parenrightBig 2 in the denominator since r 2 s 2 , and we then find that the net ycomponent of the electric field due to the left dipole is given by E y = k qs r 3 . Now, lets consider the field due to the upper dipole. The field due to the positive charge points downward, while the field due to the positive charge points upward. Here, the entire electric field due to both point charges points along our line, so we need only use the net magnitude of the electric field, k q r 2 . So, we have: E y,top = k q ( d s 2 ) 2 k q ( r + s 2 ) 2 = kq parenleftBigg ( r + s 2 ) 2 ( r s 2 ) 2 ( r 2 ( s 2 ) 2 ) 2 parenrightBigg = kq parenleftBigg r 2 + rs + 1 4 s 2 ( r 2 rs + 1 4 s 2 ) ( r 2 ( s 2 ) 2 ) 2 parenrightBigg = k 2 qrs ( r 2 ( s 2 ) 2 ) 2 And, once again, we make the approxima tion that r 2 s 2 , which leaves us with E y,top = k 2 qrs d 4 = k 2 qs r 3 Version 081 midterm 01 chiu (56565) 2 So, finally, we find that the net electric field in the y direction due to both dipoles is: E y,net = E y,left + E y,top = k qs r 3 + k 2 qs r 3 = 3 k qs r 3 = 3(9 10 9 N m 2 / C 2 ) (5 nC)(2 mm) (10 cm) 3 (1 10 9 C / nC)(0 . 001 m / mm) (0 . 01 m / cm) 3 = 270 N / C 002 10.0 points A positive charge + q sits at the location shown in the first figure, causing a field of vector E 1 A at point A . Now two identical blocks of....
View
Full
Document
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 at Austin.
 Fall '08
 Turner
 Physics

Click to edit the document details