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# 05exam1 - Husain Zeena Exam 1 Due 11:00 pm Inst Sonia Paban...

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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 μ 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 ˆ ı = - ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 56 × 10 - 6 C) 2 (1 m) 2 = - 0 . 0218721 N F 3 = - k q 2 L 2 ˆ = - ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 56 × 10 - 6 C) 2 (1 m) 2 = - 0 . 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 = - 0 . 00773296 N k ~ F k 2 = ( F 1 + F 4 x ) 2 + ( F 3 + F 4 y ) 2 = ( - 0 . 0218721 N + 0 . 00773296 N) 2 + ( - 0 . 0218721 N + - 0 . 00773296 N) 2 so that k ~ F k = 0 . 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

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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. 1. k ~ E P k = Q ² 0 A 2. k ~ E P k = 4 π ² 0 a Q 3. k ~ E P k = ² 0 Q A 4. k ~ E P k = Q 4 π ² 0 a 2 5. k ~ E P k = 2 ² 0 Q A 6. k ~ E P k = 4 π ² 0 a 2 Q 7. k ~ E P k = Q 2 ² 0 A correct 8. k ~ E P k = Q 4 π ² 0 a 9. k ~ E P k = ² 0 Q a 2 Explanation: Basic Concepts Gauss’ Law, electrostatic properties of conductors.
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05exam1 - Husain Zeena Exam 1 Due 11:00 pm Inst Sonia Paban...

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