4. E-field 2-solutions

4. E-field 2-solutions - gilvin (jg47854) 4. E-field 2...

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Unformatted text preview: 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 (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 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 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 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 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 q r 2 + parenleftbigg d 2 parenrightbigg 2 sin Since sin = d 2 bracketleftBigg r 2 + parenleftbigg d 2 parenrightbigg 2 bracketrightBigg 1 / 2 , E 1 = 1 4 q d bracketleftBigg r 2 + parenleftbigg d 2 parenrightbigg 2 bracketrightBigg 3 / 2 1 4 q d r 3 , for r d, so vector E 1 1 4 q d r 3 p 1 4 (4 . 3 10- 6 C)(1 m) (100 m) 3 p . 0386465 N / C p | E 1 | . 0386465 N / C . The cos 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 ? 1. In the direction of r . 2. In a direction perpendicular to both the dipole vector p and r . 3. In the direction opposite to that of the dipole vector; e.g. , p . correct 4. In the direction of the dipole vector; e.g. , p . 5. In a direction perpendicular to the dipole vector; e.g. , p . Explanation: The electric field goes from positive charge to negative charge. The dipole vector vectorp is shown in the figure in the question. gilvin (jg47854) 4. E-field 2 meyers (21235) 2 003 (part 3 of 4) 10.0 points What is the magnitude of the electric field at P 2 ? Correct answer: 0 . 0772929 N / C. Explanation: E 2 = q 4 1 parenleftbigg r d 2 parenrightbigg 2 1 parenleftbigg r + d 2 parenrightbigg 2 = q 4 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 4 r parenleftbigg d 2 parenrightbigg r 4 , for r d, so 1 2 q d r 3 vector E 2 1 2 q d r 3 p 1 2 (4 . 3 10- 6 C)(1 m) (100 m) 3 p 1 2 (4 . 3 10- 6 C)(1 m) (100 m) 3 p . 0772929 N / C p Thus | E 2 | . 0772929 N / C ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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4. E-field 2-solutions - gilvin (jg47854) 4. E-field 2...

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