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Unformatted text preview: howard (cah3459) homework 05 Turner (56705) 1 This printout should have 12 questions. Multiplechoice 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 . 4 5 . 5 c m . 11 C 1 g A r e a l c h a r g e d e n s i t y . 1 7 C / m 2 Find the angle the thread makes with the vertically charge sheet. Correct answer: 6 . 15029 . Explanation: Let : g = 9 . 8 m / s 2 , = 8 . 854 10 12 C 2 / N m 2 , m = 1 g = 0 . 001 kg , = 0 . 17 C / m 2 = 1 . 7 10 7 C / m 2 , q = 0 . 11 C = 1 . 1 10 7 C , and L = 45 . 5 cm = 0 . 455 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 xdirection and is E = 2 . In the and directions, force equilibrium tells us T sin = q 2 T cos = m g tan = T sin T cos = q 2 m g = arctan q 2 m g = arctan (1 . 1 10 7 C) (1 . 7 10 7 C / m 2 ) 2 (0 . 001 kg) (9 . 8 m / s 2 ) = 6 . 15029 . 002 (part 2 of 2) 10.0 points What value would in order for he angle 85 ? Correct answer: 18 . 0324 C / m 2 . Explanation: From the previous part, we know = 2 m g tan q = 2 (0 . 001 kg) ( 9 . 8 m / s 2 ) tan(85 ) 1 . 1 10 7 C (8 . 854 10 12 C 2 / N m 2 ) 10 6 C 1 C = 18 . 0324 C / m 2 . 003 10.0 points Two large, parallel, insulating plates are charged uniformly with the same positive areal charge density + , which is the charge per unit area. The permittivity of free space = 1 4 k e ....
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 Fall '08
 Turner

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