Homework 2-solutions - im(gi768 Homework 2 markert(55325...

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im (gi768) – 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. 8 . 1 m 60 5 . 9 C 5 . 9 C 5 . 9 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: 1 . 07761 × 10 9 N / C. Explanation: Let : a = 8 . 1 m , q = 5 . 9 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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im (gi768) – 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 ) ( 5 . 9 C) (8 . 1 m) 2 = 1 . 07761 × 10 9 N / C . 003(part2of2)10.0points Find the direction of the field vector vector E at P . 1. ˆ ı 2. 1 2 ı ˆ ) 3. ˆ correct 4. ˆ 5. ˆ ı 6. 1 2 ı + ˆ ) 7. 1 2 ı ˆ ) 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 . 2 C 8 . 4 C p 1 . 6 m 1 . 7 m 2 . 6 m 2 . 9 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: 9 . 70795 × 10 8 N / C. Explanation: Let : ( x p , y p ) = (0 , 0) , ( x 1 , y 1 ) = (2 . 6 m , 1 . 6 m) , ( x 2 , y 2 ) = ( 2 . 9 m , 1 . 7 m) , q 1 = 8 . 2 C , q 2 = 8 . 4 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 . 6 m 2 . 6 m vextendsingle vextendsingle vextendsingle vextendsingle = 328 . 392 , θ 2 = 180 + tan 1 vextendsingle vextendsingle vextendsingle vextendsingle 1 . 7 m 2 . 9 m vextendsingle vextendsingle vextendsingle vextendsingle = 210 . 379 . 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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im (gi768) – Homework 2 – markert – (55325) 3 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × 8 . 2 C (3 . 05287 m) 2 2 . 6 m 3 . 05287 m = 6 . 73449 × 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 · m 2 / C 2 ) × 8 . 4 C (3 . 36155 m) 2 2 . 9 m 3 . 36155 m = 5 . 7637 × 10 9 N / C , so E x = E x 1 + E x
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