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4. E-field 2-solutions

# 4. E-field 2-solutions - gilvin(jg47854 4 E-eld 2...

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gilvin (jg47854) – 4. E-field 2 – meyers – (21235) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This homework is due Friday, February 4, at midnight Tucson time. 001(part1of4)10.0points 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 . 3 μ C . + 1 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 . 0386465 N / C. Explanation: Let : r = 100 m , d = 1 m , and q = 4 . 3 μ C = 4 . 3 × 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 . 3 × 10 - 6 C)(1 m) (100 m) 3 ˆ p ≈ − 0 . 0386465 N / C ˆ p | E 1 | ≈ 0 . 0386465 N / C . The cos θ components cancel due to symme- try. 002(part2of4)10.0points What is the direction of the electric field vector E 1 at P 1 ? 1. In the direction of ˆ r . 2. In a direction perpendicular to both the dipole vector ˆ p and ˆ r . 3. In the direction opposite to that of the dipole vector; e.g. , ˆ p . correct 4. In the direction of the dipole vector; e.g. , ˆ p . 5. In a direction perpendicular to the dipole vector; e.g. , ˆ p . Explanation: The electric field goes from positive charge to negative charge. The dipole vector vectorp is shown in the figure in the question.

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gilvin (jg47854) – 4. E-field 2 – meyers – (21235) 2 003(part3of4)10.0points What is the magnitude of the electric field at P 2 ? Correct answer: 0 . 0772929 N / C. Explanation: E 2 = q 4 π ǫ 0 1 parenleftbigg r d 2 parenrightbigg 2 1 parenleftbigg r + d 2 parenrightbigg 2 = q 4 π ǫ 0 parenleftbigg r + d 2 parenrightbigg 2 parenleftbigg r d 2 parenrightbigg 2 parenleftbigg r d 2 parenrightbigg 2 parenleftbigg r + d 2 parenrightbigg 2 q 4 π ǫ 0 4 r parenleftbigg d 2 parenrightbigg r 4 , for r d , so 1 2 π ǫ 0 q d r 3 vector E 2 1 2 π ǫ 0 q d r 3 ˆ p 1 2 π ǫ 0 (4 . 3 × 10 - 6 C)(1 m) (100 m) 3 ˆ p 1 2 π ǫ 0 (4 . 3 × 10 - 6 C)(1 m) (100 m) 3 ˆ p 0 . 0772929 N / C ˆ p Thus | E 2 | ≈ 0 . 0772929 N / C . 004(part4of4)10.0points What is the direction of the electric field vector E 2 at P 2 ?
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