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Solution to UniPhysics_II_HW #4

Solution to UniPhysics_II_HW #4 - Solution to-HW...

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Solution to -- HW assignment #4 Chapter 31: Electric Current and Resistance Note: you may need to use the following simplified relations to solve the following HW problems 1) dt dQ I = , 2) E j σ = , 3) Ed V = , 4) σ ρ 1 = 5) A I j / = , 6) A L R ρ = --------------------------------------------------------------------------------------------------------------- Solve: (a) The current associated with the moving film is the rate at which the charge on the film moves past a certain point. The tangential speed of the film is ( )( ) rev 1 min 2 rad 90 rpm 4.0 cm 90 1.0 cm 9.425 cm/s min 60 s 1 rev v r π ω = = = × × × = In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is (9.425 cm)(4.0 cm) = 37.7 cm 2 . The amount of charge that passes to the right in 1.0 s is Q = (37.7 cm 2 )( 2.0 × 10 9 C/cm 2 ) = 75.4 × 10 9 C Since I Q t = Δ , we have ( ) 9 75.4 10 C 75.4 nA 1 s I × = = The current is 75 nA. (b) Having found the current in part (a), we can once again use I Q t = Δ to obtain Δ t : 6 9 10 10 C 133 s 75.4 10 A Q t I × Δ = = = × ---------------------------------------------------------------------------------------------------------------------------------- Solve: (a) The values of I (A) for selected values of t are t ( μ s) 0 1 2 4 6 8 10 I (A) 2.0 1.21 0.74 0.27 0.10 0.036 0.014 (b) Because I dQ dt = , ( ) ( ) ( ) ( ) 2.0 s 2.0 s 2.0 s 0 0 2.0 A 4.0 A s 4.0 C 1 t t t t t Q I dt e dt e e μ μ μ μ μ = = = = where we have used the condition Q = 0 C at t = 0 μ s. (c) The values of Q ( μ C) for selected values of t are t ( μ s) 0 1 2 4 6 8 10 Q ( μ C) 0 1.57 2.52 3.46 3.80 3.92 3.98
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