This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: benavides (jjb2356) oldfinal 01 Turner (59130) 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points 2 6 c m 2 6 c m 90 5 C 9 C 9 C What is the magnitude of the electrostatic force bardbl vector F bardbl on the top charge? The Coulomb constant is 8 . 9875 10 9 N m 2 / C 2 . Correct answer: 8 . 46096 N. Explanation: Let : Q = 9 C , q = 5 C , and L = 26 cm . L q Q Q bardbl vector F bardbl = k e  q 1  q 2  r 2 By symmetry and the fact that force on charge q by + Q is repulsive and by Q is attractive, F y = 0 . The x component of the forces on q by Q and Q are equal in magnitude and direction. Note: cos 45 = 2 2 . Hence, bardbl vector F bardbl = 2 k e q Q L 2 2 2 = 2 (8 . 9875 10 9 N m 2 / C 2 ) (5 C) (9 C) (26 cm) 2 2 2 = 8 . 46096 N . keywords: 002 10.0 points A circular arc has a uniform linear charge density of 9 nC / m. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 2 4 5 2 . 7 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 50 . 5335 N / C. Explanation: Let : = 9 nC / m = 9 10 9 C / m , = 245 , and r = 2 . 7 m . is defined as the angle in the counter clockwise direction from the positive x axis as shown in the figure below. 1 2 2 . 5 1 2 2 . 5 r vector E benavides (jjb2356) oldfinal 01 Turner (59130) 2 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 yaxis, 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 righthalf 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 = 90 122 . 5 = 32 . 5 , is the angle from the positive x axis to the righthand end of the arc. E = 2 k e parenleftBigg r integraldisplay 90 32 . 5 sin d parenrightBigg = 2 k e r [cos ( 32 . 5 ) cos (90 )] ....
View
Full
Document
This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

Click to edit the document details