Chap24_04 - .2 The square surface shown in Fig 24-25...

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Chapter 24 24.2 The square surface shown in Fig 24-25 measures 3.2 mm on each side. It is immersed ina 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 N m 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. 24.3. The cube in Fig 24-26 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 r A = (1.4 m ) 2 ˆ j We can now compute flux. (a) r E r A = 6.00 ˆ i m ) 2 ˆ j = 0 (b) r E r A = 2.00 ˆ j m ) 2 ˆ j = 2.00 m ) 2 = 3.92 Nm 2 / C (c) r E r 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.
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24.4 You have four point charges 2q,q,-q,-2a. If possible, describe how you would place a closes surface that enclose at least the charge 2q (and perhaps other charges) and through which the net electric flux is (a) 0, (b ) 3 q / ε 0 and (c) 2 q / 0 a) 2q q -q -2q b) 2q q -q -2q c) Not possible.
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24.9. It is found experimentally that the electric field i a certain region o Earth’s atmosphere is directed vertically down. At an altitude of 300m, the field has magnitude 60N/C; at an altitude of 200m, the magnitude is 100N/C. Find the net amount of charge contained in a cube 100m on edge with horizontal faces at altitudes of 200 and 300 m. Neglect the curvature of the earth. We calculate the net flux ϕ = 60 N / C (100 m ) 2 cos180 + 100 N / C m ) 2 cos0 = 100 N / C m ) 2 60 N / C m ) 2 = 40 N / C m ) 2 = 4 × 10 5 q enc = ε 0 = 8.85 × 10 12 4 × 10 5 = 3.54 × 10 6 C
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24.11 A point charge q is placed at one corner of a cube of edge a. What is the flux through each of the cube surfaces? To capture 1/2 the total flux, we would need 4 cubes as shown in the figure (where q is in the corner of each as shown). These 4 cubes have 3 external square sides each. So 1/2 the total flux goes through 12 identical surfaces. The flux through one square on external surface of one cube is ϕ = 1 12 closed 2 closed = q ε 0 = 1 24 q 0 24.14 Space vehicles traveling through Earth’s radiation belt can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations.
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This note was uploaded on 04/27/2008 for the course PHYS 2422 taught by Professor Harding during the Spring '08 term at University of Mary Hardin–Baylor.

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Chap24_04 - .2 The square surface shown in Fig 24-25...

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