Chapter_22_Solutions

# Chapter_22_Solutions - r r 22.1 a = E A =(14 N/C(0.250 m2...

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22.1: a) Φ= r E r A = (14 N/C) (0.250 m 2 ) cos 60 °= 1.75 Nm 2 C. b) As long as the sheet is flat, its shape does not matter. c i) The maximum flux occurs at an angle φ = 0 ° between the normal and field. c ii) The minimum flux occurs at an angle = 90 ° 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.

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22.2: a) Φ= r E r A = EAcos θ where r A = n ˆ n S 1 =− ˆ j (left) Φ S 1 (4 × 10 3 NC )(0.1 m) 2 cos(90 36.9 ° ) = − 24 N m 2 C ˆ n S 2 =+ ˆ k (top) Φ S 2 (4 × 10 3 )(0.1 m) 2 cos 90 °= 0 ˆ n S 3 ˆ j (right) Φ S 3 (4 × 10 3 )(0.1 m) 2 cos (90 °− 36.9 ° ) 24 N m 2 C ˆ n S 4 ˆ k (bottom) Φ S 4 = × 10 3 )(0.1 m) 2 cos 90 0 ˆ n S 5 ˆ i (front) Φ S 5 × 10 3 )(0.1 m) 2 cos 36.9 32 N m 2 C ˆ n S 6 ˆ i (back) Φ S 6 (4 × 10 3 )(0.1 m) 2 cos 36.9 °=− 32 N m 2 C b) The total flux through the cube must be zero; any flux entering the cube must also leave it.
22.3: a) Given that r E =− B ˆ i + C ˆ j D ˆ k±, Φ= r E r A ,edge length L , and ˆ n S 1 = − ˆ j ⇒ Φ 1 = r E A ˆ n S 1 = − CL 2 . ˆ n S 2 =+ ˆ k ⇒Φ 2 = r E n S 2 DL 2 . ˆ n S 3 ˆ j 3 = r E A ˆ n S 3 CL 2 . ˆ n S 4 ˆ k 4 = r E A ˆ n S 4 DL 2 . ˆ n S 5 ˆ i 5 = r E A ˆ n S 5 BL 2 . ˆ n S 6 ˆ i 6 = r E A ˆ n S 6 BL 2 . b) Total flux i = 0 i = 1 6

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22.4: Φ= r E r A = (75.0 N C ) (0.240 m 2 ) cos70 °= 6.16 Nm 2 C.
22.5: a) Φ= r E r A = λ 2 πε 0 r (2 π rl) = l ε 0 = 6.00 × 10 6 Cm ( ) (0.400 m) 0 = 2.71 × 10 5 Nm 2 C. 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 l = 0.800 m, the flux would increase by a factor of two: 5.42 × 10 5 Nm 2 C.

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22.6: a) Φ S 1 = q 1 ε 0 = (4.00 × 10 9 C) 0 = 452 Nm 2 C. b) Φ S 2 = q 2 0 = ( 7.80 × 10 9 C) 0 =− 881 Nm 2 C. c) Φ S 3 = (q 1 + q 2 ) 0 = ((4.00 7.80) × 10 9 C) 0 429 Nm 2 C. d) Φ S 4 = (q 1 + q 2 ) 0 = ((4.00 + 2.40) × 10 9 C) 0 = 723 Nm 2 C. e) Φ S 5 = 1 + q 2 + q 3 ) 0 = ((4.00 7.80 + 2.40) × 10 9 C) 0 158 Nm 2 C. 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) Φ= q ε 0 = ( 3.60 × 10 6 C) 0 =− 4.07 × 10 5 Nm 2 C. b) q 0 q = 0 0 (780 Nm 2 C ) = 6.90 × 10 9 C. c) No. All that matters is the total charge enclosed by the cube, not the details of where the charge is located.

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22.8: a) No charge enclosed so Φ = 0 b) Φ= q 2 ε 0 = 6.00 × 10 9 C 8.85 × 10 12 C 2 Nm 2 =− 678 Nm 2 C. c) q 1 + q 2 0 = (4.00 6.00) × 10 9 C 8.85 × 10 12 C 2 Nm 2 226 Nm 2 C.
22.9: a) Since is uniform, the flux through a closed surface must be zero. That is: r E Φ= r E d r A = q ε 0 = 1 0 ρ dV = 0 ⇒∫ dV = 0. But because we can choose any volume we want, 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.

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22.10: a) If ρ > 0 and uniform, then inside any closed surface is greater than zero. q ⇒Φ> 0 r E d r A > 0 and so the electric field cannot be uniform, i.e., since an arbitrary
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## This note was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

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Chapter_22_Solutions - r r 22.1 a = E A =(14 N/C(0.250 m2...

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