homework 03-solutions - howard (cah3459) – homework 03...

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howard (cah3459) – 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 AB 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 = 4 1 . correct 2. E A E B = 1 1 . 3. E A E B = 2 1 . 4. E A E B = 8 1 . 5. E A E B = 1 2 . 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 = 2 Q π 2. Δ q = Q Δ θ π correct 3. Δ q = π Q 4. Δ q 5. Δ q = Q 6. Δ q = Q 2 π 7. Δ q = 2 Q Δ θ π 8. None oF these 9. Δ q = Q Δ θ 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 | r 2 2. Δ E x = k | Δ q | cos θ r 3. Δ E x = k | Δ q | (cos θ ) r
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howard (cah3459) – homework 03 – Turner – (56705) 2 4. Δ E x = k | Δ q | (sin θ ) r 2 5. Δ E x = k | Δ q | (cos θ ) r 2 6. Δ E x = k | Δ q | sin θ r 2 7. Δ E x = k | Δ q | cos θ r 2 correct 8. Δ E x = k | Δ q | sin θ r 9. Δ E x = k | Δ q | r 2 10. Δ E x = k | Δ q | (sin θ ) r Explanation:
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homework 03-solutions - howard (cah3459) – homework 03...

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