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Unformatted text preview: maldonado (om2892) Homework 3 2010 UPII Stone (13750) 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 A uniformly charged insulating rod of length 16 . 3 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 1 6 . 3 c m 6 . 62 C O If the rod has a total charge of 6 . 62 C, find the horizontal component of the electric field at O , the center of the semicircle. Define right as positive. Correct answer: 1 . 40703 10 7 N / C. Explanation: Let : L = 16 . 3 cm = 0 . 163 m and q = 6 . 62 C = 6 . 62 10 6 C . Call the length of the rod L and its charge q . Due to symmetry E y = integraldisplay dE y = 0 and E x = integraldisplay dE sin = k e integraldisplay dq sin r 2 , where dq = dx = r d , so that b y x E x = k e r integraldisplay 3 / 2 / 2 cos d = k e r (sin ) vextendsingle vextendsingle vextendsingle vextendsingle 3 / 2 / 2 = 2 k e r , where = q L and r = L . Therefore, E x = 2 k e q L 2 = 2 (8 . 98755 10 9 N m 2 / C 2 ) (0 . 163 m) 2 ( 6 . 62 10 6 C) = 1 . 40703 10 7 N / C . Since the rod has negative charge, the field is pointing to the left (towards the charge dis tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 002 (part 1 of 5) 2.0 points Consider a cylindrical distribution [dark gray] extending from r = 0 to r = 9 . 7 cm, of charge = r a , where a = 30 cm m 3 / C. This cylindrical charge distribution [dark gray] is surrounded by a dielectric shell [light gray] whose inner radius is 17 cm and outer radius is 2 . 6 cm as shown in the figure. L 2 . 6 cm 17 cm 9 . 7 cm b maldonado (om2892) Homework 3 2010 UPII Stone (13750) 2 What is the electric field at 3 . 66 cm? As sume the length L is very long compared to the diameter of the dielectric shell. Neglect edge effects. Correct answer: 168 . 101 V / m. Explanation: Let : r = 3 . 66 cm . Gauss law contintegraldisplay vector E dvectora = Q enc = 1 integraldisplay r ( r ) dV where dV = d ( r 2 L ) = 2 r Ldr . In our case, Gauss law gives 2 r LE = 1 a integraldisplay r r 2 r Ldr, E = r 2 3 a = 1 8 . 85419 10 12 C 2 / N m 2 (3 . 66 cm) 2 3 (30 cm m 3 / C) 1 m 100 cm 1 C 10 6 C = 168 . 101 V / m . 003 (part 2 of 5) 2.0 points What is the electric field at 12 . 7 cm? Correct answer: 901 . 821 V / m. Explanation: Let : r = 12 . 7 cm . Now the integration in the RHS of Gauss law goes up to the boundary of our charge distribution 2 r E = 1 a integraldisplay R r 2 r dr, giving E = R 3 3 a r = 1 8 . 85419 10 12 C 2 / N m 2 (9 . 7 cm) 3 3 (30 cm m 3 / C)(12 . 7 cm) 1 m 100 cm 1 C 10 6 C = 901 . 821 V / m ....
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This note was uploaded on 03/31/2011 for the course PHYS 1001 taught by Professor Stone during the Summer '10 term at Texas Brownsville.
 Summer '10
 Stone
 Charge, Work

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