# quiz1 - Daniels Matthew – Quiz 1 – Due 11:00 pm –...

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Unformatted text preview: Daniels, Matthew – Quiz 1 – Due: Sep 21 2006, 11:00 pm – Inst: Ditmire 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The work needed to carry a 1 C charge from point A to point B is 10 J. Calculate the potential difference between point A and B. Correct answer: 10 V. Explanation: Let : q = 1 C and W = 10 J . The work is W = q V V = W q = 10 J 1 C = 10 V . keywords: 002 (part 1 of 1) 10 points A particle with charge- 4 μ C is located on the x-axis at the point- 4 cm , and a second particle with charge 8 μ C is placed on the x-axis at- 8 cm . The Coulomb constant is 8 . 9875 × 10 9 N · m 2 / C 2 . 2 4 6 8 10- 2- 4- 6- 8- 10- 4 μ C 8 μ C- 5 μ C x → (cm) What is the magnitude of the total elec- trostatic force on a third particle with charge- 5 μ C placed on the x-axis at- 2 cm? Correct answer: 349 . 514 N. Explanation: Let : q 1 =- 4 μ C =- 4 × 10- 6 C , q 2 = 8 μ C = 8 × 10- 6 C , q 3 =- 5 μ C =- 5 × 10- 6 C , x 1 =- 4 cm =- . 04 m , x 2 =- 8 cm =- . 08 m , and x 3 =- 2 cm =- . 02 m . Coulomb’s law (in vector form) for the elec- tric force exerted by a charge q 1 on a second charge q 3 , written ~ F 13 is ~ F 13 = k e q 1 q 3 r 2 ˆ r 13 , where ˆ r 13 is a unit vector directed from q 1 to q 3 ; i.e. , ~r 13 = ~r 3- ~r 1 . x 13 = x 3- x 1 = (- 2 cm)- (- 4 cm) = 0 . 02 m x 23 = x 3- x 2 = (- 2 cm)- (- 8 cm) = 0 . 06 m ˆ x 13 = x 3- x 1 p ( x 3- x 1 ) 2 = +ˆ ı ˆ x 23 = x 3- x 2 p ( x 3- x 2 ) 2 = +ˆ ı Since the forces are collinear, the force on the third particle is the algebraic sum of the forces between the first and third and the second and third particles. ~ F = ~ F 13 + ~ F 23 = k e • q 1 r 2 13 ˆ r 13 + q 2 r 2 23 ˆ r 23 ‚ q 3 = 8 . 9875 × 10 9 N · m 2 / C 2 × • (- 4 × 10- 6 C) (0 . 02 m) 2 (+ˆ ı ) + (8 × 10- 6 C) (0 . 06 m) 2 (+ˆ ı ) ‚ × (- 5 × 10- 6 C) = (449 . 375 N) + (- 99 . 8611 N) = 349 . 514 N k ~ F k = 349 . 514 N . keywords: 003 (part 1 of 1) 10 points Daniels, Matthew – Quiz 1 – Due: Sep 21 2006, 11:00 pm – Inst: Ditmire 2 A wire that has a uniform linear charge den- sity of 1 . 7 μ C / m is bent into the shape as shown below, with radius 1 . 3 m. 2 . 6 m 2 . 6 m p 1 . 3 m The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Find the electrical potential at point p . Correct answer: 81570 . 9 V. Explanation: Let : λ = 1 . 7 μ C / m = 1 . 7 × 10- 6 C / m , R = 1 . 3 m , 2 R = 2 . 6 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . 2 R 2 R p R Let p be the origin. Consider the potential due to the line of charge to the right of p ....
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## This note was uploaded on 06/23/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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quiz1 - Daniels Matthew – Quiz 1 – Due 11:00 pm –...

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