# hw 03 - nguyen (jmn727) homework 03 Turner (59070) 1 This...

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Unformatted text preview: nguyen (jmn727) homework 03 Turner (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1. E A E B = 1 2 . 2. E A E B = 2 1 . 3. E A E B = 4 1 . correct 4. E A E B = 8 1 . 5. E A E B = 1 1 . Explanation: Let : r B = 2 r A . The electric field strength E 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 002 (part 1 of 3) 10.0 points Consider the setup shown in the figure be- low, where the arc is a semicircle with radius r . The total charge Q is negative, and dis- tributed uniformly on the semicircle. The charge on a small segment with angle is labeled q . x y------------------ r x y I II III IV B A O q is given by 1. q = Q 2. q = Q 3. q = 2 Q 4. None of these 5. q = Q 2 6. q = 2 Q 7. q = Q correct 8. q = Q 2 9. q = 2 Q 10. q = Q Explanation: The angle of a semicircle is , thus the charge on a small segment with angle is q = Q . 003 (part 2 of 3) 10.0 points The magnitude of the x-component of the electric field at the center, due to q , is given by 1. E x = k | q | cos r 2. E x = k | q | r 2 3. E x = k | q | (sin ) r 2 nguyen (jmn727) homework 03 Turner (59070) 2 4. E x = k | q | r 2 5. E x = k | q | sin r 6. E x = k | q | cos r 2 correct 7. E x = k | q | (sin ) r 8. E x = k | q | (cos...
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## hw 03 - nguyen (jmn727) homework 03 Turner (59070) 1 This...

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