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Chapter 22

# Chapter 22 - 2 2 E A(14 N/C(0.250 m cos 60 1 75 Nm C 22.1 a...

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22.1: a) C. Nm 75 . 1 60 cos ) m (0.250 N/C) 14 ( 2 2 = ° = = Φ A E b) As long as the sheet is flat, its shape does not matter. ci) The maximum flux occurs at an angle ° = 0 φ between the normal and field. cii) The minimum flux occurs at an angle ° = 90 φ between the normal and field. In part i), the paper is oriented to “capture” the most field lines whereas in ii) the area is oriented so that it “captures” no field lines. 22.2: a) n A A E ˆ where cos A θ EA = = = Φ C m N 32 9 . 36 cos m) (0.1 ) C N 10 4 ( (back) ˆ ˆ C m N 32 9 . 36 cos m) (0.1 ) C N 10 4 ( front) ( ˆ ˆ 0 90 cos m) (0.1 ) C N 10 4 ( bottom) ( ˆ ˆ C m N 24 ) 9 . 36 (90 cos m) (0.1 ) C N 10 4 ( right) ( ˆ ˆ 0 90 cos m) (0.1 ) C N 10 4 ( top) ( ˆ ˆ C m N 24 ) 9 36 90 ( cos m) (0.1 ) C N 10 4 ( left) ( ˆ ˆ 2 2 3 2 2 3 2 3 2 2 3 2 3 S 2 2 3 6 6 5 5 4 4 3 3 2 2 1 1 - = ° × - = Φ - = = ° × + = Φ + = = ° × = Φ - = + = ° - ° × + = Φ + = = ° × - = Φ + = - = ° - × - = Φ - = S S S S S S S S S S S . i n i n k n j n k n j n b) The total flux through the cube must be zero; any flux entering the cube must also leave it. 22.3: a) Given that length edge , , ˆ D ˆ C ˆ B A E k j i E = Φ - + - = L, and . BL ˆ ˆ ˆ . BL ˆ ˆ ˆ . DL ˆ ˆ ˆ . CL ˆ ˆ ˆ . DL ˆ ˆ ˆ . CL ˆ ˆ ˆ 2 6 2 5 2 4 2 3 2 2 2 1 6 6 5 5 4 4 3 3 2 2 1 1 + = = Φ - = - = = Φ + = + = = Φ - = + = = Φ + = - = = Φ + = - = = Φ - = S S S S S S S S S S S S A A A A A A n E i n n E i n n E k n n E j n n E k n n E j n b) Total flux = = Φ = 6 1 i 0 i 22.4: C. Nm 16 . 6 70 cos ) m (0.240 ) C N 0 . 75 ( 2 2 = ° = = Φ A E

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22.5: a) C. Nm 10 71 . 2 ) 2 ( 2 5 m) (0.400 C/m) 10 00 . 6 ( 2 0 6 0 0 × = = = = = Φ - × λ λ ε ε l r πε πrl A E b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the field lines must still pass through the surface. c) If the length was increased to m, 800 . 0 = l the flux would increase by a factor of two: C. Nm 10 5.42 2 5 × = Φ 22.6: a) C. Nm 452 C) 10 00 . 4 ( 2 0 9 0 1 1 = × = = Φ - ε ε q S b) C. Nm 881 C) 10 80 . 7 ( 2 0 9 0 2 2 - = × - = = Φ - ε ε q S c) C. Nm 429 C) 10 ) 80 . 7 00 . 4 ( ( ) ( 2 0 9 0 2 1 3 - = × - = + = Φ - ε ε q q S d) C. Nm 723 C) 10 ) 40 . 2 00 . 4 ( ( ) ( 2 0 9 0 2 1 4 = × + = + = Φ - ε ε q q S e) C. Nm 158 C) 10 ) 40 . 2 80 . 7 00 . 4 ( ( ) ( 2 0 9 0 3 2 1 5 - = × + - = + + = Φ - ε ε q q q S f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface. 22.7: a) C. Nm 10 07 . 4 C) 10 60 . 3 ( 2 5 0 6 0 × - = × - = = Φ - ε ε q b) C. 10 6.90 ) C Nm 780 ( 9 2 0 0 0 - × = = Φ = = Φ ε ε q ε q c) No. All that matters is the total charge enclosed by the cube, not the details of where the charge is located. 22.8: a) No charge enclosed so 0 = Φ b) C. Nm 678 Nm C 10 8.85 C 10 00 . 6 2 2 2 12 9 0 2 - = × × - = = Φ - - ε q c) C. Nm 226 Nm C 10 8.85 C 10 ) 00 . 6 00 . 4 ( 2 2 2 12 9 0 2 1 - = × × - = + = Φ - - ε q q 22.9: a) Since E is uniform, the flux through a closed surface must be zero. That is: = = = = = Φ . 0 0 0 0 1 ρdV ρdV d ε ε q A E But because we can choose any volume we want, ρ must be zero if the integral equals zero. b) If there is no charge in a region of space, that does NOT mean that the electric field is uniform. Consider a closed volume close to, but not including, a point charge. The field diverges there, but there is no charge in that region.
22.10: a) If 0 ρ and uniform, then q inside any closed surface is greater than zero.

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