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651_PartUniversity Physics Solution

# 651_PartUniversity Physics Solution - 22-2 Chapter...

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22-2 Chapter 22 (b) Total flux: 2 2 2 5 (0.081 0.135)(N/C) m 0.054 N m /C. Φ = Φ + Φ = = − Therefore, 13 0 4.78 10 C. q = Φ = − × P E VALUATE : Flux is positive when E G is directed out of the volume and negative when it is directed into the volume. 22.5. I DENTIFY : 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. S ET U P : The electric field is perpendicular to the flat circle, so the flux is simply the product of E and the area of the flat circle of radius r . E XECUTE : Φ E = EA = E ( 2 r π ) = 2 r π E E VALUATE : The flux would be the same if the hemisphere were replaced by any other surface bounded by the flat circle. 22.6. I DENTIFY : Use Eq.(22.3) to calculate the flux for each surface. S ET U P : ˆ cos where EA A φ Φ = = E A A = n G G G . E XECUTE : (a) 1 ˆ ˆ (left) S n = j . 1 3 2 2 (4 10 N/C)(0.10 m) cos(90 36 9 ) 24 N m /C. S . Φ = − × ° = − ° 2 S ˆ ˆ (top) + n = k . 2 3 2 (4 10 N/C)(0.10 m) cos90 0 S Φ = − × ° = . 3 ˆ ˆ (right) S + n = j . 3 3 2 2 (4 10 N/C)(0.10 m) cos(90 36.9 ) 24 N m /C S Φ = + × ° − ° = + . 4 ˆ ˆ (bottom) S n = k . 4 3 2 (4 10 N/C)(0.10 m) cos90 0 S Φ = × ° = . 5 ˆ ˆ (front) S + n = i . 5 3 2 2 (4 10 N/C)(0.10 m) cos36.9 32 N m /C S Φ = + × ° = . 6 ˆ ˆ (back) S n = i . 6 3 2 2 (4 10 N/C)(0.10 m) cos36.9 32 N m /C S Φ = − × ° = − . E VALUATE : (b) The total flux through the cube must be zero; any flux entering the cube must also leave it, since the field is uniform. Our calculation gives the result; the sum of the fluxes calculated in part (a) is zero. 22.7. (a) I DENTIFY : Use Eq.(22.5) to calculate the flux through the surface of the cylinder. S ET U P : The line of charge and the cylinder are sketched in Figure 22.7. Figure 22.7 E XECUTE : The area of the curved part of the cylinder is 2 . A rl π = The electric field is parallel to the end caps of the cylinder, so 0 = E A G G for the ends and the flux through the cylinder end caps is zero. The electric field is normal to the curved surface of the cylinder and has the same magnitude 0 /2 E r λ π = P at all points on this surface. Thus 0 φ = ° and ( )( ) ( ) ( ) 6 5 2 0 12 2 2 0 6.00 10 C/m 0.400 m cos / 2 2 2.71 10 N m /C 8.854 10 C / N m E l EA EA r rl λ φ λ π π × Φ = = = = = = × × P P (b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same, 5 2 2.71 10 N m /C. E Φ = × (c) ( ) ( ) 6 5 2 12 2 2 0 6.00 10 C/m 0.800 m 5.42 10 N m /C 8.854 10 C / N m E l λ × Φ = = = × × P (twice the flux calculated in parts (b) and (c)). E VALUATE : The flux depends on the number of field lines that pass through the surface of the cylinder. 22.8. I DENTIFY : Apply Gauss’s law to each surface. S ET U P : encl Q is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and flux into the enclosed volume is negative.

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