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fall_06_exam4_sol_phy2049

fall_06_exam4_sol_phy2049 - PHY2049 Fall 2006 Prof Yasu...

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PHY2049 Fall 2006 1 Prof. Yasu Takano Prof. Paul Avery Dec. 11, 2006 Exam 4 Solutions ( First answer is correct ) 1. ( Exam 1 ) Consider a cylinder of radius 6 m and height 15 m. At each point on the cylindrical surface, an electric field of 10 V/m points out radially. There is no electric field on the top and bottom surfaces of the cylinder. What is the charge enclosed by the cylinder in coulombs? (1) 5.0 × 10 8 (2) 2.5 × 10 8 (3) 1.7 × 10 8 (4) 8.3 × 10 9 (5) 3.3 × 10 8 We use Gauss’ law, and note that only the E field perpendicular to the curved surface contrib- utes. Using the area of a cylinder, Gauss’ law yields ( ) enc 0 2 / E rh q π ε = , where r is the cylinder radius and h is its height. Solving for the enclosed charge yields 8 enc 5.0 10 q = × C. 2. ( WebAssign 24.5 ) Two large, parallel, conducting plates are 1.2 mm apart and have equal and opposite charges on their facing surfaces. An electrostatic force of 4 × 10 13 N acts on an elec- tron placed anywhere between the two plates. What is the potential difference in volts between the plates? (1) 3000 (2) 360 (3) 1.88 × 10 6 (4) 1880 (5) 2250 The force on an electron is F = eE. The potential difference between two plates is Ed. Thus the potential difference can be written V = Fd/e = 3000 V.
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