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# HW 3 - benavides(jjb2356 homework 03 Turner(59130 y This...

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benavides (jjb2356) – homework 03 – Turner – (59130) 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 0 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 = 2 1 . 2. E A E B = 4 1 . correct 3. E A E B = 1 1 . 4. E A E B = 1 2 . 5. E A E B = 8 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 Δ θ is labeled Δ q . x y - - - - - - - - - - - - - - - - - - Δ θ θ r x y I II III IV B A O Δ q is given by 1. Δ q = Q π 2. Δ q = Q Δ θ 2 π 3. Δ q = Q 2 π 4. Δ q = 2 Q π 5. Δ q = 2 π Q 6. Δ q = Q Δ θ π correct 7. None of these 8. Δ q = π Q 9. Δ q = 2 Q Δ θ π 10. Δ q = Q Explanation: The angle of a semicircle is π , thus the charge on a small segment with angle Δ θ is Δ q = Q Δ θ π . 003 (part 2 of 3) 10.0 points The magnitude of the x -component of the electric field at the center, due to Δ q , is given by 1. Δ E x = k | Δ q | (cos θ ) r 2. Δ E x = k | Δ q | cos θ r 3. Δ E x = k | Δ q | (sin θ ) r

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benavides (jjb2356) – homework 03 – Turner – (59130) 2 4. Δ E x = k | Δ q | sin θ r 2 5. Δ E x = k | Δ q | (sin θ ) r 2 6. Δ E x = k | Δ q | r 2 7. Δ E x = k | Δ q | r 2 8. Δ E x = k | Δ q | (cos θ ) r 2 9. Δ E x = k | Δ q | sin θ r 10. Δ E x = k | Δ q | cos θ r 2 correct Explanation: Negative charge attracts a positive test charge. At O , Δ E
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