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hw2 solutions

# hw2 solutions - homework 02 BOBBITT JAMES Due 4:00 am...

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homework 02 – BOBBITT, JAMES – Due: Jan 28 2008, 4:00 am 1 Question 1, chap 23, sect 2. part 1 of 3 10 points A uniformly charged circular arc AB of ra- dius R is shown in the figure. It covers a quarter of a circle and it is located in the sec- ond quadrant. The total charge on the arc is Q > 0. The value of the Coloumb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . x y + + + + + + + + + Δ θ θ R x y I II III IV B A O Δ s R Δ θ What is the direction of the electric field vector vector E at the origin, due to the charge distribution? 1. along the negative x -axis 2. in quadrant II 3. along the positive y -axis 4. in quadrant I 5. along the positive x -axis 6. along the negative y -axis 7. in quadrant IV correct 8. in quadrant III Explanation: The electric field for a positive charge is directed away from it. In this case, the electric field generated by each Δ q will be directed into quadrant IV, so the total electric field will be in the same quadrant. Question 2, chap 23, sect 2. part 2 of 3 10 points Find E x , the x -component of the electric field at the origin due to the full arc length for the case, where Q = 2 . 6 μ C and R = 0 . 48 m. Correct answer: 64567 . 3 N / C (tolerance ± 1 %). Explanation: Let : Q = 2 . 6 μ C = 2 . 6 × 10 6 C , R = 0 . 48 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . E x can be found by integrating the contri- butions of all the Δ q ’s in the arc. Using Δ q = Δ θ parenleftBig π 2 parenrightBig and taking the x -component, then adding the force component from each charge Δ q at angle θ , we have E x = integraldisplay π/ 2 0 k e Q R 2 2 π cos θ dθ = 2 k e Q π R 2 = 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) π (0 . 48 m) 2 × (2 . 6 × 10 6 C) = 64567 . 3 N / C . Question 3, chap 23, sect 2. part 3 of 3 10 points If at O we have E x = C , then what is the magnitude of the full force F (not just the x -component) on an electron at this point? 1. F = C e 2 2. F = 2 C 3 e 3. F = 2 e C 4. F = 2 e C correct 5. F = 3 e C 2 6. F = 4 e C

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homework 02 – BOBBITT, JAMES – Due: Jan 28 2008, 4:00 am 2 7. F = C 2 e 8. F = e C 9. F = C 3 e 10. F = 3 e C Explanation: From the symmetry of the charge distri- bution, it can be seen that E y = E x = C , so E x E y E E = radicalBig E 2 x + E 2 y = 2 C . The force on an electron is then F = e E = 2 e C . Question 4, chap 23, sect 2. part 1 of 4 10 points Consider a disk of radius 4 . 6 cm, having a uniformly distributed charge of +6 . 8 μ C. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Compute the magnitude of the electric field at a point on the axis and 3 . 4 mm from the center. Correct answer: 5 . 3507 × 10 7 N / C (tolerance ± 1 %). Explanation: Let : R = 4 . 6 cm = 0 . 046 m , k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 6 . 8 μ C = 6 . 8 × 10 6 C , and x = 3 . 4 mm = 0 . 0034 m . The field at the distance x along the axis of a disk with radius R is given by E = 2 π k e σ parenleftbigg 1 x x 2 + R 2 parenrightbigg , where the surface charge density σ is σ = Q π R 2 = 6 . 8 × 10 6 C π (0 . 046 m) 2 = 0 . 00102292 C / m 2 .
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