Chap23_06 - Chapter 23 Problems 23.1 The square surface...

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Chapter 23 Problems 23.1 The square surface shown in Fig 23-26 measures 3.2 mm on each side. It is immersed in a uniform electric field with magnitude E = 1800 N/C. The field lines make an angle of 35 degrees with a normal to the surface as shown. Take the normal to be directed “outward” as though the surface were one face of a box. Calculate the electric flux through the surface. The flux through this surface is ϕ = E A cos θ = 180 ° 35 ° = (1800 N / C ) (.0032 m ) 2 cos(180 ° 35 ° ) = 1.51 × 10 2 Nm 2 / C Note that the angle is 180-35. This makes the flux negative--which means the flow is into the box. A net flow into a closed surface is taken to be negative. 23.2. The cube in Fig 23-27 has edge length of 1.4 m and is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the field (in N/C) is given by (a) 6.00 ˆ i , (b) 2.00 ˆ j and (c) 3.00 ˆ i + 4.00 ˆ k . The area vector for the right face is A = (1.4 m ) 2 ˆ j We can now compute flux. (a) E A = 6.00 ˆ i m ) 2 ˆ j = 0 (b) E A = 2.00 ˆ j m ) 2 ˆ j = 2.00 m ) 2 = 3.92 Nm 2 / C (c) E A = ( 3.00 ˆ i + 4.00 ˆ k ) m ) 2 ˆ j = 0 (d) The total flux through the cube is zero. A uniform field is present--every field line that enters onside of the cube leaves the other. 23.4 In Fig. 23-28 a butterfly net is in a uniform electric field of magnitude E = 3.0 mN / C . The rim, a circle of radius a = 11.0 cm is aligned perpendicular to the filed. The net contains no net charge. Find the electric flux through the netting The netting plus the circle make a closed surface. We can use this surface and Gauss’ Law to find the flux through the netting. We begin with Gauss
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E d A = q enc ε 0 E d A netting + E d A circle = q enc 0 E d A netting + E d A circle = 0 E d A netting = E d A circle E d A netting = 3 mN / C π (0.11 m ) 2 = 0.114 mN C m 2 23.7 In Fig. 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). If we think of the charge as enclosed by a cube with the charge at the center, we can use Gauss’ Law to find the flux through the cube ϕ cube = q enc 0 = 1.81 × 10 8 Cm 2 side = 1 6 cube = 3.01 × 10 9 N C m 2 23.8 When a shower is turned on in a closed bathroom, the splashing of the water on the bare tube can fill the room’s air with negatively charged ions and produce an electric field in the air as great as 1000 N/C. Consider a bathroom with dimension 2.5m x 3.0 m x 2.0m . Along the ceiling floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of 600 N/C. Also, treat those surfaces as forming a closed Gaussian surface around the room’s air. What are (a) the volume charge density ρ and (b)the number of excess elementary charges e per cubic meter.
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This note was uploaded on 05/22/2008 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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Chap23_06 - Chapter 23 Problems 23.1 The square surface...

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