22_InstructorSolutions

22_InstructorSolutions - GAUSS'S LAW 22 ^ E = E cos dA...

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22-1 G AUSS S L AW 22.1. (a) I DENTIFY and S ET U P : cos , E E dA φ Φ = where φ is the angle between the normal to the sheet ˆ n and the electric field . E G E XECUTE : In this problem E and cos φ are constant over the surface so ( )( ) ( ) 2 2 cos cos 14 N/C cos60 0.250 m 1.8 N m /C. E E dA E A φ φ Φ = = = ° = (b) E VALUATE : E Φ is independent of the shape of the sheet as long as φ and E are constant at all points on the sheet. (c) E XECUTE : (i) cos . E E E A φ Φ = Φ is largest for 0 , so cos 1 and . E EA φ φ = ° = Φ = (ii) E Φ is smallest for 90 , so cos 0 and 0. E φ φ = ° = Φ = E VALUATE : E Φ is 0 when the surface is parallel to the field so no electric field lines pass through the surface. 22.2. I DENTIFY : The field is uniform and the surface is flat, so use cos E EA φ Φ = . S ET U P : φ is the angle between the normal to the surface and the direction of E G , so 70 φ = ° . E XECUTE : 2 (75.0 N/C)(0.400 m)(0.600 m)cos70 6.16 N m /C E Φ = = ° E VALUATE : If the field were perpendicular to the surface the flux would be 2 18.0 N m /C. E EA Φ = = 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. I DENTIFY : 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) S ET U P : In this case, the electric field is perpendicular to the surface of the sphere, so 2 (4 ) E EA E r π Φ = = . E XECUTE : Substituting in the numbers gives ( ) ( ) 2 6 5 2 1.25 10 N/C 4 0.150 m 3.53 10 N m /C E π Φ = × = × (b) I DENTIFY : We use the electric field due to a point charge. S ET U P : 2 0 1 4 q E r π = P E XECUTE : Solving for q and substituting the numbers gives ( ) ( ) 2 2 6 6 0 9 2 2 1 4 0.150 m 1.25 10 N/C 3.13 10 C 9.00 10 N m /C q r E π = = × = × × P E VALUATE : 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. I DENTIFY : Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge. S ET U P : ˆ ˆ ( 5.00 N/C m) (3.00 N/C m) x z E = i + k G . The area of each face is 2 L , where 0.300 m L = . E XECUTE : 1 1 1 ˆ ˆ ˆ 0 s S A ⇒ Φ = = n = j E n G . 2 2 2 2 ˆ ˆ ˆ (3.00 N C m)(0.300 m) (0.27 (N C) m) S S A z z + ⇒ Φ = = = n = k E n G . 2 2 (0.27 (N/C)m)(0.300 m) 0.081 (N/C) m Φ = = . 3 3 3 ˆ ˆ ˆ 0 S S A + ⇒ Φ = = n = j E n G . 4 4 4 ˆ ˆ ˆ (0.27 (N/C) m) 0 (since 0). S S A z z ⇒ Φ = = − = = n = k E n G 5 5 2 5 ˆ ˆ ˆ ( 5.00 N/C m)(0.300 m) (0.45 (N/C) m) . S S A x x + ⇒ Φ = = − = − n = i E n G 2 5 (0.45 (N/C) m)(0.300 m) (0.135 (N/C) m ). Φ = − = − 6 6 6 ˆ ˆ ˆ (0.45 (N/C) m) 0 (since 0). S S A x x − ⇒ Φ = = + = = n = i E n G 22

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