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# solution_pdf5 - tovar(jdt436 homework 05 Turner(59070 This...

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tovar (jdt436) – homework 05 – Turner – (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet of infinite extent. The acceleration of gravity is 9 . 8 m / s 2 and the permittivity of free space is 8 . 854 × 10 12 C 2 / N · m 2 . 31 . 3 cm θ ˆ ı ˆ 0 . 21 μ C 1 g Areal charge density 0 . 12 μ C / m 2 Find the angle θ the thread makes with the vertically charge sheet. Correct answer: 8 . 26233 . Explanation: Let : g = 9 . 8 m / s 2 , ǫ 0 = 8 . 854 × 10 12 C 2 / N · m 2 , m = 1 g = 0 . 001 kg , σ = 0 . 12 μ C / m 2 = 1 . 2 × 10 7 C / m 2 , q = 0 . 21 μ C = 2 . 1 × 10 7 C , and L = 31 . 3 cm = 0 . 313 m . The length L of the string is superfluous. Let the tension in the string be denoted by T . The electric field due to the infinite sheet is constant in the x -direction and is vector E = σ 2 ǫ 0 ˆ ı . In the ˆ ı and ˆ directions, force equilibrium tells us T sin θ = q · σ 2 ǫ 0 T cos θ = m g tan θ = T sin θ T cos θ = q σ 2 m g ǫ 0 θ = arctan parenleftbigg q σ 2 m g ǫ 0 parenrightbigg = arctan bracketleftbigg (2 . 1 × 10 7 C) (1 . 2 × 10 7 C / m 2 ) 2 (0 . 001 kg) (9 . 8 m / s 2 ) ǫ 0 bracketrightbigg = 8 . 26233 . 002 (part 2 of 2) 10.0 points What value would σ in order for he angle 85 ? Correct answer: 9 . 44556 μ C / m 2 . Explanation: From the previous part, we know σ = 2 m g tan θ q ǫ 0 = 2 (0 . 001 kg) ( 9 . 8 m / s 2 ) tan(85 ) 2 . 1 × 10 7 C · (8 . 854 × 10 12 C 2 / N · m 2 ) 10 6 μ C 1 C = 9 . 44556 μ C / m 2 . 003 10.0 points A net positive charge Q is placed on a large, thin conducting plate of are A . In electrostatic equilibrium the charge den- sity σ on each surface and the electric field E outside the plate are 1. σ = 0 , E = σ ǫ 0 2. σ = Q A , E = σ ǫ 0 3. σ = Q 2 A , E = σ ǫ 0 correct 4. σ = Q 2 A , E = 0 5. σ = Q 2 A , E = σ 2 ǫ 0

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tovar (jdt436) – homework 05 – Turner – (59070) 2 Explanation: 004 10.0 points Two large, parallel, insulating plates are charged uniformly with the same positive areal charge density + σ , which is the charge per unit area. The permittivity of free space ǫ 0 = 1 4 π k e .
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solution_pdf5 - tovar(jdt436 homework 05 Turner(59070 This...

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