HW2-Solutions - Version PREVIEW HW 2 feng (58230) This...

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Version PREVIEW – HW 2 – feng – (58230) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. Uniformly Charged Bent Wire 01 001 10.0 points A uniformly charged insulating rod of length 13 . 2 cm is bent into the shape of a semicircle as in the Fgure. The value of the Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . 1 3 . 2 c m 7 . 48 μ C O If the rod has a total charge of 7 . 48 μ C, Fnd the horizontal component of the electric Feld at O , the center of the semicircle. DeFne right as positive. Correct answer: 2 . 42424 × 10 7 N / C. Explanation: Let : L = 13 . 2 cm = 0 . 132 m and q = 7 . 48 μ C = 7 . 48 × 10 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = i dE y = 0 and E x = i dE sin θ = k e i dq sin θ r 2 , where dq = λ dx = λ r dθ , so that b y x θ E x = k e λ r i 3 π/ 2 π/ 2 cos θ dθ = k e λ r (sin θ ) v v v v 3 π/ 2 π/ 2 = 2 k e λ r , where λ = q L and r = L π . Therefore, E x = 2 k e q π L 2 = 2 (8 . 98755 × 10 9 N m 2 / C 2 ) (0 . 132 m) 2 × ( 7 . 48 × 10 6 C) π = 2 . 42424 × 10 7 N / C . Since the rod has negative charge, the Feld is pointing to the left (towards the charge dis- tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. Charged Arc 03 002 (part 1 of 2) 10.0 points A uniformly charged circular arc AB of radius R is shown in the Fgure. It covers a quarter of a circle and it is located in the second quadrant. The total charge on the arc is Q > 0. The value of the Coulomb constant is 8 . 99 × 10 9 N · m 2 / C 2 .
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Version PREVIEW – HW 2 – feng – (58230) 2 x y + + + + + + + + + Δ θ θ R x y I II III IV B A O Δ s R Δ θ The direction of the electric Feld vector v E at the origin, due to the charge distribution, is 1. along the negative y -axis 2. along the negative x -axis 3. along the positive y -axis 4. in quadrant III 5. in quadrant II 6. in quadrant I 7. in quadrant IV correct 8. along the positive x -axis Explanation: Basic Concept: Electric Felds generated by charge distributions. Solution: The electric Feld for a positive charge is directed away from it. In this case, the electric Feld generated by each Δ q will be directed into quadrant IV, so the total electric Feld will be in the same quadrant. 003 (part 2 of 2) 10.0 points ±ind E x , the x -component of the electric Feld at the origin due to the full arc length for the case, where Q = 0 . 7 μ C and R = 1 . 59 m. Correct answer: 1584 . 26 N / C. Explanation: Let : Q = 0 . 7 μ C and R = 1 . 59 m . E x can be found by integrating the con- tributions of all the Δ q ’s in the arc. Us- ing Δ q = Q π/ 2 Δ θ and then adding the x- component of the Feld from each charge ele- ment Δ q at angle θ , we have i π/ 2 0 k e Q R 2 2 π cos θ dθ = 2 k e Q π R 2 = 2 k e (0 . 7 μ C) π (1 . 59 m) 2 = 1584 .
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HW2-Solutions - Version PREVIEW HW 2 feng (58230) This...

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