Fisica halliday resolucao

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 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.

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: