hw 04 - nguyen (jmn727) homework 04 Turner (59070) This...

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nguyen (jmn727) – homework 04 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 3) 10.0 points A rod oF length with uniForm charge per unit length λ , where λ > 0, is placed a distance d From the origin along the x axis. A similar rod with the same charge is placed along the y axis as in the fgure. Consider only a small segment oF length Δ x (it is oF infnitesimal length) on the horizontal rod, which is placed along the x axis. The segment is at a distance x From the origin. x y d λ O d λ Determine the electric feld Δ v E at the ori- gin due to this segment. 1. Δ v E = k e λ Δ x x ˆ ı 2. Δ v E = k e λ Δ x x ˆ ı 3. Δ v E = k e λ Δ x 2 x ˆ ı 4. Δ v E = k e λ Δ x 2 x ˆ ı 5. Δ v E = k e λ Δ x x 2 ˆ ı correct 6. Δ v E = k e λ Δ x x 2 ˆ ı Explanation: ±rom the Fact that λ > 0 and the diagram, the feld Δ E at O due to the segment on the x -axis is in the negative x direction, and its magnitude is given by Δ E = k e Δ q x 2 = k e λ Δ x x 2 . We can thus express the electric feld as Δ v E = Δ E ˆ ı . 002 (part 2 oF 3) 10.0 points Determine the electric feld v E rod 1 at the origin due to the entire horizontal rod. 1. v E rod 1 = k e λ d ˆ ı 2. v E rod 1 = k e λ ℓ d 2 ˆ ı 3. v E rod 1 = k e λ d ˆ ı 4. v E rod 1 = k e λ ℓ d ˆ ı 5. v E rod 1 = k e λ ℓ d ˆ ı 6. v E rod 1 = k e λ d ( + d ) ˆ ı 7. v E rod 1 = k e λ d ( + d ) ˆ ı 8. v E rod 1 = k e λ ℓ d ( d + ) ˆ ı correct 9. v E rod 1 = k e λ d 2 ˆ ı 10. v E rod 1 = k e λ ℓ d ( d + ) ˆ ı Explanation: ±or the entire rod, we must add all the segments, i.e. , integrate From x = d to x = d + . The electric feld due to the horizontal rod is v E rod 1 = i d v E = k
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hw 04 - nguyen (jmn727) homework 04 Turner (59070) This...

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