HW3 - yang (ey942) oldhomework 03 Turner (56705) 1 This...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: yang (ey942) oldhomework 03 Turner (56705) 1 This print-out should have 14 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 8 nC / m. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 2 3 1 2 . 5 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 51 . 917 N / C. Explanation: Let : = 8 nC / m = 8 10 9 C / m , = 231 , and r = 2 . 5 m . is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 1 1 5 . 5 1 1 5 . 5 r vector E 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 r In polar coordinates dq = ( r d ) , where is the linear charge density. The positive y axis is = 90 , so the y component of the electric field is given by dE y = dE sin . 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 = 90 . The lower angular limit = 90 115 . 5 = 25 . 5 , is the angle from the positive x axis to the right-hand end of the arc. E = 2 k e parenleftBigg r integraldisplay 90 25 . 5 sin d parenrightBigg = 2 k e r [cos ( 25 . 5 ) cos (90 )] . Since k e r = (8 . 98755 10 9 N m 2 / C 2 ) ( 8 10 9 C / m) (2 . 5 m) = 28 . 7602 N / C , E = 2 ( 28 . 7602 N / C) [(0 . 902585) (0)] = 51 . 917 N / C bardbl vector E bardbl = 51 . 917 N / C . yang (ey942) oldhomework 03 Turner (56705) 2 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). is still defined as the angle in the counter- clockwise direction from the positive x axis. E x = parenleftBigg k e r integraldisplay 231 cos d parenrightBigg = k e r [sin (231 ) sin(0 )] = ( 28 . 7602 N / C) [( . 777147) . 0] = [ 22 . 3509 N / C] , E y = parenleftBigg k e r integraldisplay 231 sin d parenrightBigg = k e r [cos (0 ) cos (231 )] = ( 28 . 7602 N / C) [1 . ( . 629319)] = [46 . 8595 N / C] , bardbl...
View Full Document

Page1 / 8

HW3 - yang (ey942) oldhomework 03 Turner (56705) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online