303L Midterm01 - Version 118 midterm01 berk(60290 This...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 118 – midterm01 – berk – (60290) 1 This print-out should have 15 questions. Multiple-choice 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 . 278 2 . 8 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? 1. 9.3248 2. 18.4109 3. 24.7485 4. 46.9173 5. 29.3779 6. 25.5445 7. 22.101 8. 25.2701 9. 37.9847 10. 42.2536 Correct answer: 25 . 2701 N / C. Explanation: Let : λ = 6 nC / m = 6 × 10 9 C / m , Δ θ = 278 , and r = 2 . 8 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 139 139 r vector 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 y -axis, so E x = 0 . For a continuous linear charge distribution, vector E = k e integraldisplay 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 right-half 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 = 90 139 = 49 , is the angle from the positive x axis to the right-hand end of the arc. E = 2 k e parenleftBigg λ r integraldisplay 90 49 sin θ dθ parenrightBigg ˆ = 2 k e λ r [cos ( 49 ) cos (90 )] ˆ  . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( 6 × 10 9 C / m) (2 . 8 m) = 19 . 259 N / C ,
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 118 – midterm01 – berk – (60290) 2 E = 2 ( 19 . 259 N / C) × [(0 . 656059) (0)] ˆ = 25 . 2701 N / C ˆ bardbl vector E bardbl = 25 . 2701 N / C . Alternate Solution: Just solve for bardbl vector E bardbl 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 = parenleftBigg k e λ r integraldisplay 278 0 cos θ dθ parenrightBigg ˆ ı = k e λ r [sin (278 ) sin (0 )] ˆ ı = ( 19 . 259 N / C) × [( 0 . 990268) 0 . 0] ˆ ı = [ 19 . 0716 N / C] ˆ ı , E y = parenleftBigg k e λ r integraldisplay 278 0 sin θ dθ parenrightBigg ˆ = k e λ r [cos (0 ) cos (278 )] ˆ = ( 19 . 259 N / C) × [1 . 0 (0 . 139175)] ˆ = [16 . 5787 N / C] ˆ  , bardbl vector E bardbl = radicalBig E 2 x + E 2 y = bracketleftBig ( 19 . 0716 N / C) 2 + (16 . 5787 N / C) 2 bracketrightBig 1 / 2 = 25 . 2701 N / C .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern