ass6 - SOLUTIONS, ASSIGNMENT 6 22.3. IDENTIFY:The electric...

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SOLUTIONS, ASSIGNMENT 6 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 ( 29 ( 29 2 6 5 2 1.25 10 N/C 4 0.150 m 3.53 10 N m /C E 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 = P E XECUTE : Solving for q and substituting the numbers gives ( 29 ( 29 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 - = = × = × × 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 r . The area of each face is 2 L , where 0.300 m L = . E
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ass6 - SOLUTIONS, ASSIGNMENT 6 22.3. IDENTIFY:The electric...

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