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HW2 - oliver(jmo673 – HW02 – Tsoi –(59090 1 This...

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Unformatted text preview: oliver (jmo673) – HW02 – Tsoi – (59090) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find 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 10 cm 20 cm The ratio of the electric field strength at point A to the electric field 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 = 8 1 . 4. E A E B = 1 2 . 5. E A E B = 2 1 . Explanation: Let : r B = 2 r A . The electric field 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 2) 10.0 points A charge Q is distributed uniformly along the x axis from x 1 to x 2 . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 Which of the following integrals is correct for the magnitude of the electric field at x on the x axis? Assume that x > x 2 > x 1 and k e = 1 4 πǫ . 1. None of these 2. integraldisplay x 2 x 1 k e Q ( x 2 − x )( x − x ) 2 dx 3. integraldisplay x 2 x 1 k e Q ( x 2 − x ) x 2 dx 4. integraldisplay x 2 x 1 k e Q ( x 2 − x 1 ) x 2 dx 5. integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx correct Explanation: For a continuous charge distribution, dE = k e dq r 2 . Recall that for a uniform linear charge dis- tribution, dq = λdx = Q L dx = Q x 2 − x 1 dx . Furthermore, the point x is a distance ( x − x ) from a charge element dq . Hence integrating from x = x 1 to x = x 2 yields E = integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx . 003 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 4 . 48 m and the charge Q = 3 . 07 μ C, what is the magnitude E of the electric field at x = 9 . 9 m? Correct answer: 514 . 216 N / C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , x 1 = 0 m , x 2 = 4 . 48 m , x = 9 . 9 m , and Q = 3 . 07 μ C . oliver (jmo673) – HW02 – Tsoi – (59090) 2 E = integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx = k e Q ( x 2 − x 1 )( x − x ) vextendsingle vextendsingle vextendsingle vextendsingle x 2 x 1 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) (4 . 48 m − 0 m) × (3 . 07 × 10 − 6 C) × bracketleftbigg 1 9 . 9 m − 4 . 48 m − 1 9 . 9 m − 0 m bracketrightbigg = 514 . 216 N / C . 004 (part 1 of 3) 10.0 points Consider the setup shown in the figure 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 = π Q 2. Δ q = Q Δ θ 2 π 3. Δ q = Q π 4. Δ q = Q 2 π 5. Δ q = 2 Q Δ θ π 6. None of these 7. Δ q = 2 π Q 8. Δ q = Q Δ θ π correct 9. Δ q = 2 Q π 10. Δ q = Q Explanation: The angle of a semicircle is π , thus the charge on a small segment with angle Δ θ is Δ...
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HW2 - oliver(jmo673 – HW02 – Tsoi –(59090 1 This...

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