# Momento inercia

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Física III – Sears, Zemansky, Young & Freedman.

PHYSICS ACT.
http//physicsact.wordpress.com

Capítulo 22

r r 22.1: a) Φ = E ⋅ 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. ci) The maximum flux occurs at an angle φ = 0° between the normal and field. cii) The minimum flux occurs at an angle φ = 90° between the normal andfield. 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.

r r r ˆ 22.2: a) Φ = E ⋅ A = EA cos θ where A = An ˆ nS1 = − ˆ (left)Φ S1 = −(4 × 10 3 N C) (0.1 m) 2 cos (90 − 36.9°) = −24 N ⋅ m 2 C j

ˆ ˆ nS2 = + k ( top)Φ S2 = −(4 × 103 N C) (0.1 m) 2 cos 90° = 0 ˆ nS3 = + ˆ (right)Φ S3 = +(4 × 103 N C) (0.1 m)2 cos (90° − 36.9°) = +24 N ⋅ m 2 C j ˆ ˆ nS4 = −k (bottom)Φ S4 = (4 × 103 N C) (0.1 m) 2 cos 90° = 0 ˆ ˆ nS5 = + i (front)Φ S5 = + (4 × 10 3 N C) (0.1 m) 2 cos 36.9° = 32 N ⋅ m 2 C ˆ ˆ nS6 = −i (back)Φ S6 = −(4 × 103 N C) (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.
r r r ˆ ˆ j 22.3: a) Given that E = − Bi +Cˆ − Dk , Φ = E ⋅ A, edge length L, and r ˆ ˆ nS1 = − ˆ ⇒ Φ1 = E ⋅ AnS1 = −CL2 . j r ˆ ˆ ˆ nS2 = + k ⇒ Φ 2 = E ⋅ AnS2 = −DL2 . r ˆ ˆ nS3 = + ˆ ⇒ Φ 3 = E ⋅ AnS3 = +CL2 . j r ˆ ˆ ˆ nS4 = −k ⇒ Φ 4 = E ⋅ AnS4 = + DL2 . r ˆ ˆ ˆ nS5 = + i ⇒ Φ 5 = E ⋅ AnS5 = −BL2 . r ˆ ˆ ˆ nS6 = −i ⇒ Φ 6 = E ⋅ AnS6 = + BL2 .

b) Total flux = ∑i =1 Φ i = 0
6

22.4:

r r Φ = E ⋅ A = (75.0 N C) (0.240 m 2 ) cos 70° =6.16 Nm 2 C.

r r 22.5: a) Φ = E ⋅ A =

λ 2 πε0 r

(2πrl ) =

λl ε0

=

( 6.00×10−6 C/m) (0.400 m) ε0

= 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: Φ = 5.42 × 105 Nm 2 C.

22.6:a) Φ S1 = q1 ε0 = (4.00 × 10 −9 C) ε 0 = 452 Nm 2 C.

b) Φ S2 = q2 ε0 = (−7.80 × 10 −9 C) ε0 = −881 Nm 2 C.
c) Φ S3 = (q1 + q2 ) ε 0 = ((4.00 − 7.80) × 10 −9 C) ε 0 = −429 Nm 2 C. d) Φ S4 = (q1 + q2 ) ε0 = ((4.00 + 2.40) × 10 −9 C) ε0 = 723 Nm 2 C. e) Φ S5 = (q1 + q2 + q3 ) ε0 = ((4.00 − 7.80 + 2.40) × 10 −9 C) ε0 = −158 Nm 2 C. f) All that matters for Gauss’s law is the total amount of chargeenclosed by the surface, not its distribution within the surface.

22.7: a) Φ = q ε0 = (−3.60 × 10 −6 C) ε0 = −4.07 × 105 Nm 2 C. b) Φ = q ε0 ⇒ q = ε0 Φ = ε0 (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.
22.8: a) No charge enclosed so Φ = 0 q − 6.00 × 10 −9 C b) Φ= 2 = = − 678 Nm 2 C. ε0 8.85 ×10 −12 C 2 Nm 2 c)
Φ=

q1 + q2 (4.00 − 6.00) × 10 −9 C = = −226 Nm 2 C. ε0 8.85 × 10 −12 C 2 Nm 2

r 22.9: a) Since E is uniform, the flux through a closed surface must be zero. That is: r r Φ = ∫ E ⋅ dA = εq0 = ε10 ∫ ρdV = 0 ⇒ ∫ ρdV = 0. 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 NOTmean 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 ρ > 0 and uniform, then q inside any closed surface is greater than zero. r r ⇒ Φ > 0 ⇒ ∫ E ⋅ dA > 0 and so the electric field cannot be uniform, i.e., since an arbitrary surface of our choice encloses anon-zero amount of charge, E must depend on position. b) However, inside a small bubble of zero density within the material with density ρ , the field CAN be uniform. All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge). (See Exercise 22.61.) 22.11: Φ 6sides = q ε0 = (9.60 × 10 −6 C) ε0 = 1.08 × 10 6 Nm 2 C. But the box is symmetrical, so for...