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Unformatted text preview: Husain, Zeena Exam 1 Due: Feb 17 2004, 11:00 pm Inst: Sonia Paban 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Four point charges are placed at the four cor ners of a square. Each side of the square has a length L . q 1 = q q 3 = q q 2 = q q 4 = q P L L Find the magnitude of the electric force on q 2 due to all three charges q 1 , q 3 and q 4 . Given L = 1 m and q = 1 . 56 C. Correct answer: 0 . 0328082 N. Explanation: F 1 F 4y F 4 F 3 F 4x q 2 From the above figure, we see that F 1 = k q 2 L 2 = ( 8 . 98755 10 9 N m 2 / C 2 ) (1 . 56 10 6 C) 2 (1 m) 2 = . 0218721 N F 3 = k q 2 L 2 = ( 8 . 98755 10 9 N m 2 / C 2 ) (1 . 56 10 6 C) 2 (1 m) 2 = . 0218721 N F 4 x = k q 2 2 L 2 1 2 = 8 . 98755 10 9 N m 2 / C 2 2 (1 m) 2 (1 . 56 10 6 C) 2 2 = 0 . 00773296 N F 4 y = k q 2 2 L 2 1 2 = 8 . 98755 10 9 N m 2 / C 2 2 (1 m) 2 (1 . 56 10 6 C) 2 2 = . 00773296 N k ~ F k 2 = ( F 1 + F 4 x ) 2 + ( F 3 + F 4 y ) 2 = ( . 0218721 N + 0 . 00773296 N) 2 + ( . 0218721 N + . 00773296 N) 2 so that k ~ F k = . 00107638 N 2 = 0 . 0328082 N 002 (part 1 of 1) 10 points Consider two SOLID conducting spheres with radii r 1 and r 2 , where r 2 = 3 r 1 . The two spheres are separated by a large distance so that the field and the potential at the surface of sphere #1 only depend on the charge on #1 and the corresponding quantities on #2 only depend on the charge on #2. There is a wire connecting the two spheres. r 1 r 2 #2 #1 Place a charge q on sphere #1. After equi librium has been reached, what is the ratio E 2 E 1 of the electric fields at the SURFACES of the two solid spheres? 1. 2. 27 3. 9 Husain, Zeena Exam 1 Due: Feb 17 2004, 11:00 pm Inst: Sonia Paban 2 4. 1 81 5. 81 6. 1 27 7. 1 3 correct 8. 3 9. 1 9 10. 1 Explanation: When the spheres are connected by a wire, charges will flow from one to the other until the potential on both spheres is the same. In this case, k q 1 r 1 = k q 2 r 2 , or q 2 = r 2 r 1 q 1 . We know outside the sphere E ( r ) = k q r 2 . Therefore, E 2 E 1 = q 2 q 1 ( r 1 r 2 ) 2 = r 2 r 1 ( r 1 r 2 ) 2 = 3 1 ( 1 3 ) 2 = 1 3 003 (part 1 of 2) 10 points A uniformly charged conducting plate with area A has a total charge Q which is positive. The figure below shows a crosssectional view of the plane and the electric field lines due to the charge on the plane. The figure is not drawn to scale. E E + Q P Find the magnitude of the field at point P, which is a distance a from the plate. Assume that a is very small when compared to the dimensions of the plate, such that edge effects can be ignored....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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