Phys121-3sol

Phys121-3sol - Anderson, Matthew Homework 3 Due: Feb 12...

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Unformatted text preview: Anderson, Matthew Homework 3 Due: Feb 12 2008, noon Inst: Vitaly 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points A good model for a radio antenna is a dipole. In the figure below P 1 is on the perpendic- ular bisector of the dipole, and P 2 is along the axis of the dipole in the direction of the dipole vector p . P 1 and P 2 are both a distance 100 m from the center of the dipole. The magnitude of each of the charges is 3 . 9 C . +- 1 . 9 m ~p P 2 (- 100 m , 0) P 1 (0 , 100 m) What is the magnitude of the electric field at P 1 ? Hint: The direction of the dipole vector p is from the negative charge to the positive charge. Correct answer: 0 . 0665978 N / C. Explanation: Let : r = 100 m , d = 1 . 9 m , and q = 3 . 9 C = 3 . 9 10- 6 C . +- ~p P 2 P 1 E + E- E + E- The electric field E 1 is E 1 = 2 1 4 q r 2 + d 2 2 sin Since sin = d 2 " r 2 + d 2 2 # 1 / 2 , E 1 = 1 4 q d " r 2 + d 2 2 # 3 / 2 1 4 q d r 3 , for r d, so ~ E 1 - 1 4 q d r 3 p - 1 4 (3 . 9 10- 6 C)(1 . 9 m) (100 m) 3 p - . 0665978 N / C p | E 1 | . 0665978 N / C . The cos components cancel due to symme- try. 002 (part 2 of 4) 10 points What is the direction of the electric field ~ E 1 at P 1 ? 1. In the direction of r . 2. In a direction perpendicular to both the dipole vector p and r . 3. In a direction perpendicular to the dipole vector; e.g. , p . 4. In the direction opposite to that of the dipole vector; e.g. ,- p . correct 5. In the direction of the dipole vector; e.g. , p . Explanation: The electric field goes from positive charge to negative charge. The dipole vector ~p is shown in the figure in the question. 003 (part 3 of 4) 10 points Anderson, Matthew Homework 3 Due: Feb 12 2008, noon Inst: Vitaly 2 What is the magnitude of the electric field at P 2 ? Correct answer: 0 . 133196 N / C. Explanation: E 2 = q 4 1 r- d 2 2- 1 r + d 2 2 = q 4 r + d 2 2- r- d 2 2 r- d 2 2 r + d 2 2 q 4 4 r d 2 r 4 , for r d, so 1 2 q d r 3 ~ E 2 1 2 q d r 3 p 1 2 (3 . 9 10- 6 C)(1 . 9 m) (100 m) 3 p 1 2 (3 . 9 10- 6 C)(1 . 9 m) (100 m) 3 p . 133196 N / C p Thus | E 2 | . 133196 N / C . 004 (part 4 of 4) 10 points What is the direction of the electric field ~ E 2 at P 2 ?...
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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Phys121-3sol - Anderson, Matthew Homework 3 Due: Feb 12...

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