HW_4_solutions

HW_4_solutions - HW 4 Solutions (due online 1 AM, Friday,...

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HW 4 Solutions (due online 1 AM, Friday, January 28, 2011) 1. Chapter 23 Problem 1 The square surface shown in Fig. 23-26 measures 3.1 mm on each side. It is immersed in a uniform electric field with magnitude E = 2500 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Figure 23-26 Solutions: The vector area A and the electric field E are shown on the diagram below. The angle between them is 180° 35° = 145°, so the electric flux through the area is    2 3 2 2 cos 1800 N C 3.2 10 m cos145 1.5 10 N m C. E A EA       2. Chapter 23 Problem 5 In Figure 23-29, a proton is a distance d /2 directly above the center of a square of side d . What is the magnitude of the electric flux through the square? ( Hint : Think of the square as one face of a cube with edge d .)
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Figure 23-29
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HW_4_solutions - HW 4 Solutions (due online 1 AM, Friday,...

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