solution_pdf4 - tovar(jdt436 – homework 04 – Turner...

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Unformatted text preview: tovar (jdt436) – homework 04 – Turner – (59070) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A nonconducting plate with infinite dimen- sions carries a uniform surface charge density of 5 . 31 μ C / cm 2 . What is the electric field 9 . 8 cm in front of the plate? Correct answer: 2 . 99858 × 10 9 N / C. Explanation: Let : σ = 5 . 31 μ C / cm 2 and r = 9 . 8 cm . The electric field of an infinite plane of surface charge density σ is E = σ 2 ǫ = 5 . 31 μ C / cm 2 2 (8 . 85419 × 10 − 12 C 2 / N · m 2 ) × parenleftbigg 1 C 10 6 μ C parenrightbigg · parenleftbigg 100 cm 1 m parenrightbigg 2 = 2 . 99858 × 10 9 N / C . 002 10.0 points A ring of radius 12 cm that lies in the yz plane carries positive charge of 5 μ C uniformly distributed over its length. A particle of mass m that carries a charge of − 5 μ C oscillates about the center of the ring with an angular frequency of 24 rad / s. Find the angular frequency of oscillation of the mass if the radius of the ring is doubled to 24 cm and all other parameters remain unchanged. Correct answer: 8 . 48528 rad / s. Explanation: Let : R = 12 cm , R ′ = 24 cm , Q = 5 μ C , Q p = 5 μ C , and ω i = 24 rad / s . The electric field along the x axis is E = k Q x parenleftBig √ R 2 + x 2 parenrightBig 3 . If x ≪ R , the electric field can be approxi- mated as E = k Q x R 3 , so the force on the mass is F = k Q Q p x R 3 . Applying simple harmonic oscillation, we have ω = radicalbigg k Q Q p mR 3 or ω ∝ radicalbigg 1 R 3 , so ω ′ ω i = radicalbigg R 3 R ′ 3 = parenleftbigg R R ′ parenrightbigg 3 / 2 ω i = parenleftbigg 12 cm 24 cm parenrightbigg 3 / 2 (24 rad / s) = 8 . 48528 rad / s ....
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This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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solution_pdf4 - tovar(jdt436 – homework 04 – Turner...

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