9 - Copy (2)

9 - Copy (2) - tang (nmt352) homework 03 Turner (56705) y...

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tang (nmt352) – homework 03 – Turner – (56705) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd 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 feld strength at point A to the electric feld strength at point B is 1. E A E B = 1 1 . 2. E A E B = 2 1 . 3. E A E B = 4 1 . correct 4. E A E B = 1 2 . 5. E A E B = 8 1 . Explanation: Let : r B = 2 r A . The electric feld 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 fgure 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 Δ θ π correct 2. Δ q = 2 π Q 3. None oF these 4. Δ q = 2 Q π 5. Δ q = π Q 6. Δ q = Q Δ θ 2 π 7. Δ q = Q 8. Δ q = 2 Q Δ θ π 9. Δ q = Q 2 π 10. Δ q = Q π 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 feld 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 | (sin θ ) r 3. Δ E x = k | Δ q | sin θ r 2
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tang (nmt352) – homework 03 – Turner – (56705) 2 4. Δ E x = k | Δ q | r 2 5. Δ E x = k | Δ q | (cos θ ) r 6. Δ E x = k | Δ q | (cos θ ) r 2 7. Δ E x = k | Δ q | (sin θ ) r 2 8. Δ E x = k | Δ q | r 2 9. Δ E x = k | Δ q | cos θ r 2 correct 10. Δ E x = k | Δ q | sin θ r Explanation: Negative charge attracts a positive test charge. At
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9 - Copy (2) - tang (nmt352) homework 03 Turner (56705) y...

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