05exam1 - Husain, Zeena Exam 1 Due: Feb 17 2004, 11:00 pm...

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Unformatted text preview: Husain, Zeena Exam 1 Due: Feb 17 2004, 11:00 pm Inst: Sonia Paban 1 This print-out should have 16 questions. Multiple-choice 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 cross-sectional 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.

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05exam1 - Husain, Zeena Exam 1 Due: Feb 17 2004, 11:00 pm...

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