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Unformatted text preview: howard (cah3459) – oldhomework 03 – Turner – (56705) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circular arc has a uniform linear charge density of 6 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 2 2 ◦ 2 . 4 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 42 . 2277 N / C. Explanation: Let : λ = 6 nC / m = 6 × 10 9 C / m , Δ θ = 220 ◦ , and r = 2 . 4 m . θ is defined as the angle in the counter clockwise direction from the positive x axis as shown in the figure below. 1 1 ◦ 1 1 ◦ r # E θ First, position the arc symmetrically around the y axis, centered at the origin. By symmetry (in this rotated configuration) the field in the x direction cancels due to charge from opposites sides of the yaxis, so E x = 0 . For a continuous linear charge distribution, # E = k e dq r 2 ˆ r In polar coordinates dq = λ ( r dθ ) , where λ is the linear charge density. The positive y axis is θ = 90 ◦ , so the y component of the electric field is given by dE y = dE sin θ . Note: By symmetry, each half of the arc about the y axis contributes equally to the electric field at the origin. Hence, we may just consider the righthalf of the arc (beginning on the positive y axis and extending towards the positive x axis) and multiply the answer by 2. Note: The upper angular limit θ = 90 ◦ . The lower angular limit θ = 90 ◦ 110 ◦ = 20 ◦ , is the angle from the positive x axis to the righthand end of the arc. E = 2 k e λ r 90 ◦ 20 ◦ sin θ dθ ˆ = 2 k e λ r [cos ( 20 ◦ ) cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 6 × 10 9 C / m) (2 . 4 m) = 22 . 4689 N / C , E = 2 ( 22 . 4689 N / C) × [(0 . 939693) (0)] ˆ = 42 . 2277 N / C ˆ # E = 42 . 2277 N / C . howard (cah3459) – oldhomework 03 – Turner – (56705) 2 Alternate Solution: Just solve for # E in a straight forward manner, positioning the beginning of the arc on the positive x axis (as shown in the original figure in the question). θ is still defined as the angle in the counter clockwise direction from the positive x axis. E x = k e λ r 220 ◦ ◦ cos θ dθ ˆ ı = k e λ r [sin (220 ◦ ) sin(0 ◦ )] ˆ ı = ( 22 . 4689 N / C) × [( . 642789) . 0] ˆ ı = [ 14 . 4427 N / C] ˆ ı, E y = k e λ r 220 ◦ ◦ sin θ dθ ˆ = k e λ r [cos (0 ◦ ) cos (220 ◦ )] ˆ = ( 22 . 4689 N / C) × [1 . ( . 766043)] ˆ = [39 . 681 N / C] ˆ , # E = E 2 x + E 2 y = ( 14 . 4427 N / C) 2 + (39 . 681 N / C) 2 1 / 2 = 42 . 2277 N / C ....
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 Fall '08
 Turner
 Electrostatics, Electric charge, ex, Howard

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