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old hw 4 - Copy

# old hw 4 - Copy - vo(ytv59 oldhomework 04 Turner(56705 This...

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vo (ytv59) – oldhomework 04 – Turner – (56705) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 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 ˆ 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 4 . 8 μ C . + 1 . 6 m vectorp 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 ˆ p is from the negative charge to the positive charge. Correct answer: 0 . 0690244 N / C. Explanation: Let : r = 100 m , d = 1 . 6 m , and q = 4 . 8 μ C = 4 . 8 × 10 6 C . + vectorp P 2 P 1 E + E E + E The electric field E 1 is E 1 = 2 1 4 π ǫ 0 q r 2 + parenleftbigg d 2 parenrightbigg 2 sin θ Since sin θ = d 2 bracketleftBigg r 2 + parenleftbigg d 2 parenrightbigg 2 bracketrightBigg 1 / 2 , E 1 = 1 4 π ǫ 0 q d bracketleftBigg r 2 + parenleftbigg d 2 parenrightbigg 2 bracketrightBigg 3 / 2 1 4 π ǫ 0 q d r 3 , for r d , so vector E 1 ≈ − 1 4 π ǫ 0 q d r 3 ˆ p ≈ − 1 4 π ǫ 0 (4 . 8 × 10 6 C)(1 . 6 m) (100 m) 3 ˆ p ≈ − 0 . 0690244 N / C ˆ p | E 1 | ≈ 0 . 0690244 N / C . The cos θ components cancel due to symme- try. 002 (part 2 of 4) 10.0 points What is the direction of the electric field vector E 1 at P 1 ?

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