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Phys121-3sol

Phys121-3sol - Anderson Matthew Homework 3 Due noon Inst...

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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 – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 4) 10 points A good model ±or a radio antenna is a dipole. In the fgure below P 1 is on the perpendic- ular bisector o± the dipole, and P 2 is along the axis o± the dipole in the direction o± the dipole vector ˆ p . P 1 and P 2 are both a distance 100 m ±rom the center o± the dipole. The magnitude o± each o± the charges is 3 . 9 μ C . + - 1 . 9 m ~p P 2 ( - 100 m , 0) P 1 (0 , 100 m) What is the magnitude o± the electric feld at P 1 ? Hint: The direction o± the dipole vector ˆ p is ±rom 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 μ C = 3 . 9 × 10 - 6 C . + - P 2 P 1 E + - E + E - The electric feld E 1 is E 1 = 2 1 4 π ² 0 q r 2 + µ d 2 2 sin θ Since sin θ = d 2 " r 2 + µ d 2 2 # 1 / 2 , E 1 = 1 4 π ² 0 q d " r 2 + µ d 2 2 # 3 / 2 1 4 π ² 0 q d r 3 , ±or r À d, so ~ E 1 ≈- 1 4 π ² 0 q d r 3 ˆ p 1 4 π ² 0 (3 . 9 × 10 - 6 C)(1 . 9 m) (100 m) 3 ˆ p 0 . 0665978 N / C ˆ p | E 1 |≈ 0 . 0665978 N / C . The cos θ components cancel due to symme- try. 002 (part 2 o± 4) 10 points What is the direction o± the electric feld ~ E 1 at P 1 ? 1. In the direction o± ˆ r . 2. In a direction perpendicular to both the dipole vector ˆ p and ˆ r . 3. In a direction perpendicular to the dipole vector; e.g. , ˆ p . 4. In the direction opposite to that o± the dipole vector; e.g. , - ˆ p . correct 5. In the direction o± the dipole vector; e.g. , ˆ p . Explanation: The electric feld goes ±rom positive charge to negative charge. The dipole vector is shown in the fgure in the question. 003 (part 3 o± 4) 10 points

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Anderson, Matthew – Homework 3 – Due: Feb 12 2008, noon – Inst: Vitaly 2 What is the magnitude of the electric ±eld at P 2 ? Correct answer: 0 . 133196 N / C. Explanation: E 2 = q 4 π ² 0 1 µ r - d 2 2 - 1 µ r + d 2 2 = q 4 π ² 0 µ r + d 2 2 - µ r - d 2 2 µ r - d 2 2 µ r + d 2 2 q 4 π ² 0 4 r µ d 2 r 4 , for r À d, so 1 2 π ² 0 q d r 3 ~ E 2 1 2 π ² 0 q d r 3 ˆ p 1 2 π ² 0 (3 . 9 × 10 - 6 C)(1 . 9 m) (100 m) 3 ˆ p 1 2 π ² 0 (3 . 9 × 10 - 6 C)(1 . 9 m) (100 m) 3 ˆ p 0 . 133196 N / C ˆ p Thus | E 2 |≈ 0 . 133196 N / C . 004 (part 4 of 4) 10 points What is the direction of the electric ±eld ~ E 2 at P 2 ? 1. In a direction perpendicular to ˆ r . 2. In a direction perpendicular to the dipole vector; e.g. , ˆ p .
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Phys121-3sol - Anderson Matthew Homework 3 Due noon Inst...

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