PHY_solution-chapter22

# PHY_solution-chapter22 - 22.1 a C Nm 75 1 60 cos m(0.250...

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Unformatted text preview: 22.1: a) C. Nm 75 . 1 60 cos ) m (0.250 N/C) 14 ( 2 2 = ° = ⋅ = Φ A E As long as the sheet is flat, its shape does not matter. b) The maximum flux occurs at an angle ci) ° = φ between the normal and field. The minimum flux occurs at an angle cii) ° = 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 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 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 The total flux through the cube must be zero; any flux b) entering the cube must also leave it. 22.3: Given that a) 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 Total flux b) ∑ = = Φ = 6 1 i i 22.4: C. Nm 16 . 6 70 cos ) m (0.240 ) C N . 75 ( 2 2 = ° = ⋅ = Φ A E 22.5: a) C. Nm 10 71 . 2 ) 2 ( 2 5 m) (0.400 C/m) 10 00 . 6 ( 2 6 × = = = = ⋅ = Φ- × λ λ ε ε l r πε πrl A E We would get the same flux as in (a) if the cylinder’s b) radius was made larger—the field lines must still pass through the surface. 2613b9cf403cc776ac279c1517e0463fc51ebbf0.doc Page 1 of 23 If the length was increased to c) m, 800 . = 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 9 1 1 = × = = Φ- ε ε q S b) C. Nm 881 C) 10 80 . 7 ( 2 9 2 2- = ×- = = Φ- ε ε q S c) C. Nm 429 C) 10 ) 80 . 7 00 . 4 ( ( ) ( 2 9 2 1 3- = ×- = + = Φ- ε ε q q S d) C. Nm 723 C) 10 ) 40 . 2 00 . 4 ( ( ) ( 2 9 2 1 4 = × + = + = Φ- ε ε q q S e) C. Nm 158 C) 10 ) 40 . 2 80 . 7 00 . 4 ( ( ) ( 2 9 3 2 1 5- = × +- = + + = Φ- ε ε q q q S All that matters for Gauss’s law is the total amount of f) charge enclosed by the surface, not its distribution within the surface....
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PHY_solution-chapter22 - 22.1 a C Nm 75 1 60 cos m(0.250...

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