Homework 2-solutions

Homework 2-solutions - kumar(vk4433 Homework 2...

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kumar (vk4433) – Homework 2 – markert – (55325) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two small spheres carry electric charges of equal magnitudes. There are equally spaced points ( a , b , and c ) which lie along the same line. a b c What is the direction of the net electric field at each point due to these charges? 1. a b c 2. a b c 3. a b c 4. a b c correct 5. a b c 6. a b c 7. a b c 8. a b c 9. a b c 10. a b c Explanation: Since the field originates from positive charges and terminates on the negative charges, a b c 002(part1of2)10.0points Three point charges are placed at the vertices of an equilateral triangle. 1 . 7 m 60 0 . 2 C 0 . 2 C 0 . 2 C P ˆ ı ˆ Find the magnitude of the electric field vec- tor bardbl vector E bardbl at P . The value of the Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . Correct answer: 8 . 29296 × 10 8 N / C. Explanation: Let : a = 1 . 7 m , q = 0 . 2 C , and k = 8 . 9875 × 10 9 N · m 2 / C 2 . a q q q P ˆ ı ˆ Electric field vectors due to the bottom two charges cancel each other. h = a cos 30 =
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kumar (vk4433) – Homework 2 – markert – (55325) 2 3 2 a is the height of the triangle, so the magnitude of the field vector due to the charge at the top of the triangle is bardbl vector E bardbl = k q parenleftBigg 3 2 a parenrightBigg 2 = 4 3 k q a 2 = 4 3 (8 . 9875 × 10 9 N · m 2 / C 2 ) ( 0 . 2 C) (1 . 7 m) 2 = 8 . 29296 × 10 8 N / C . 003(part2of2)10.0points Find the direction of the field vector vector E at P . 1. 1 2 ı ˆ ) 2. ˆ 3. ˆ correct 4. ˆ ı 5. 1 2 ı + ˆ ) 6. 1 2 ı + ˆ ) 7. ˆ ı 8. 1 2 ı ˆ ) Explanation: By inspection, vector E at P is along the ˆ direc- tion. 004 10.0points Two charges are located in the ( x, y ) plane as shown. The fields produced by these charges are observed at a point p with coordinates (0 , 0). 8 . 1 C 8 C p 1 . 8 m 1 . 6 m 3 m 2 m Find the x -component of the electric field at p . The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 3 . 45859 × 10 9 N / C. Explanation: Let : ( x p , y p ) = (0 , 0) , ( x 1 , y 1 ) = (3 m , 1 . 8 m) , ( x 2 , y 2 ) = ( 2 m , 1 . 6 m) , q 1 = 8 . 1 C , q 2 = 8 C , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 q 2 p y 1 y 2 x 1 x 2 Consider the electric field vectors: θ 1 θ 2 E 1 E 2 q 1 q 2 where θ 1 = 180 tan 1 vextendsingle vextendsingle vextendsingle vextendsingle 1 . 8 m 3 m vextendsingle vextendsingle vextendsingle vextendsingle = 329 . 036 , θ 2 = 180 + tan 1 vextendsingle vextendsingle vextendsingle vextendsingle 1 . 6 m 2 m vextendsingle vextendsingle vextendsingle vextendsingle = 218 . 66 . In the x -direction, the contributions from the two charges are E x 1 = k e Q 1 r 2 1 cos θ 1 = k e Q 1 r 2 1 x 1 r 1
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kumar (vk4433) – Homework 2 – markert – (55325) 3 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × 8 . 1 C (3 . 49857 m) 2 3 m 3 . 49857 m = 5 . 10006 × 10 9 N / C and E x 2 = k e Q 2 r 2 2 cos θ 2 = k e Q 2 r 2 2 x 2 r 2 = ( 8 . 98755 × 10 9 N ·
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