YF_ISM_22

# YF_ISM_22 - GAUSS'S LAW 22 ^ E = E cos dA where is the...

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22-1 G AUSS S L AW 22.1. (a) IDENTIFY and SET UP: cos , E Ed A φ Φ= where is the angle between the normal to the sheet ˆ n and the electric field . E G EXECUTE: In this problem E and cos are constant over the surface so () 22 cos cos 14 N/C cos60 0.250 m 1.8 N m /C. E A E A φφ = = ° = (b) EVALUATE: E Φ is independent of the shape of the sheet as long as and E are constant at all points on the sheet. (c) EXECUTE: (i) cos . EE EA Φ is largest for 0, so cos 1 and . E EA = (ii) E Φ is smallest for 90 , so cos 0 and 0. E = EVALUATE: E Φ is 0 when the surface is parallel to the field so no electric field lines pass through the surface. 22.2. IDENTIFY: The field is uniform and the surface is flat, so use cos E EA . SET UP: is the angle between the normal to the surface and the direction of E G , so 70 = ° . EXECUTE: 2 (75.0 N/C)(0.400 m)(0.600 m)cos70 6.16 N m /C E = ° EVALUATE: If the field were perpendicular to the surface the flux would be 2 18.0 N m /C. E EA = The flux in this problem is much less than this because only the component of E G perpendicular to the surface contributes to the flux. 22.3. IDENTIFY: The electric flux through an area is defined as the product of the component of the electric field perpendicular to the area times the area. (a) SET UP: In this case, the electric field is perpendicular to the surface of the sphere, so 2 (4 ) E EA E r π = . EXECUTE: Substituting in the numbers gives 2 65 2 1.25 10 N/C 4 0.150 m 3.53 10 N m /C E × = × (b) IDENTIFY: We use the electric field due to a point charge. SET UP: 2 0 1 4 q E r = P EXECUTE: Solving for q and substituting the numbers gives 2 26 6 0 92 2 1 4 0.150 m 1.25 10 N/C 3.13 10 C 9.00 10 N m /C qr E == × = × ×⋅ P EVALUATE: The flux would be the same no matter how large the circle, since the area is proportional to r 2 while the electric field is proportional to 1/ r 2 . 22.4. IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge. SET UP: ˆˆ ( 5.00 N/C m) (3.00 N/C m) xz −⋅ E =i + k G . The area of each face is 2 L , where 0.300 m L = . EXECUTE: 11 1 ˆ 0 sS A = n= j En G . 2 2 ˆ (3.00 N C m)(0.300 m) (0.27 (N C) m) SS Az z + Φ= ⋅ = = n= k G . 2 2 (0.27 (N/C)m)(0.300 m) 0.081 (N/C) m = . 33 3 ˆ 0 A + = G . 44 4 ˆ (0.27 (N/C) m) 0 (since 0). z =− = = n=k G 55 2 5 ˆ ( 5.00 N/C m)(0.300 m) (0.45 (N/C) m) . Ax x + n=i G 2 5 (0.45 (N/C) m)(0.300 m) (0.135 (N/C) m ). Φ=− 66 6 ˆ (0.45 (N/C) m) 0 (since 0). x =+ = = G 22

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22-2 Chapter 22 (b) Total flux: 22 25 (0.081 0.135)(N/C) m 0.054 N m /C. Φ=Φ +Φ = =− Therefore, 13 0 4.78 10 C. q = × P EVALUATE: Flux is positive when E G is directed out of the volume and negative when it is directed into the volume. 22.5. IDENTIFY: The flux through the curved upper half of the hemisphere is the same as the flux through the flat circle defined by the bottom of the hemisphere because every electric field line that passes through the flat circle also must pass through the curved surface of the hemisphere.
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YF_ISM_22 - GAUSS'S LAW 22 ^ E = E cos dA where is the...

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